Indices and Surds Questions in English

(A) Given expression: $\left( \frac{x^l}{x^m} \right)^{l^2 + lm + m^2} \left( \frac{x^m}{x^n} \right)^{m^2 + nm + n^2} \left( \frac{x^n}{x^l} \right)^{n^2 + nl + l^2}$
Using the law of exponents $\frac{x^a}{x^b} = x^{a-b}$,we get:
$= (x^{l-m})^{(l^2 + lm + m^2)} (x^{m-n})^{(m^2 + nm + n^2)} (x^{n-l})^{(n^2 + nl + l^2)}$
Using the identity $(a-b)(a^2 + ab + b^2) = a^3 - b^3$:
$= x^{l^3 - m^3} \cdot x^{m^3 - n^3} \cdot x^{n^3 - l^3}$
Using the law of exponents $x^a \cdot x^b = x^{a+b}$:
$= x^{(l^3 - m^3 + m^3 - n^3 + n^3 - l^3)}$
$= x^0 = 1$

(D) Given $2^x = 4^y = 8^z$.
Expressing all in base $2$: $2^x = 2^{2y} = 2^{3z}$.
This implies $x = 2y = 3z = k$ (let $k$ be a constant).
Then $x = k$,$y = k/2$,and $z = k/3$.
Given $xyz = 288$,so $k \times (k/2) \times (k/3) = 288$.
$\frac{k^3}{6} = 288 \implies k^3 = 1728 \implies k = 12$.
Thus,$x = 12$,$y = 6$,and $z = 4$.
Now,calculate $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{1}{2(12)} + \frac{1}{4(6)} + \frac{1}{8(4)} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32}$.
$= \frac{2}{24} + \frac{1}{32} = \frac{1}{12} + \frac{1}{32} = \frac{8 + 3}{96} = \frac{11}{96}$.

(A) Given expression: $\frac{2 \cdot 3^{n+1} + 7 \cdot 3^{n-1}}{3^{n+2} - 2 \cdot (\frac{1}{3})^{1-n}}$
Rewrite the terms in the numerator and denominator using $3^{n-1}$ as a common factor:
Numerator: $2 \cdot 3^{n+1} + 7 \cdot 3^{n-1} = 2 \cdot 3^{n-1} \cdot 3^2 + 7 \cdot 3^{n-1} = 3^{n-1}(18 + 7) = 25 \cdot 3^{n-1}$
Denominator: $3^{n+2} - 2 \cdot (3^{-1})^{1-n} = 3^{n+2} - 2 \cdot 3^{n-1} = 3^{n-1} \cdot 3^3 - 2 \cdot 3^{n-1} = 3^{n-1}(27 - 2) = 25 \cdot 3^{n-1}$
Dividing the numerator by the denominator:
$\frac{25 \cdot 3^{n-1}}{25 \cdot 3^{n-1}} = 1$

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, \text{ and } \sqrt[6]{17}$,we express them with a common index.
The indices are $3, 4, \text{ and } 6$. The $L.C.M.$ of $3, 4, 6$ is $12$.
Converting each to the $12^{th}$ root:
$\sqrt[3]{9} = 9^{1/3} = (9^4)^{1/12} = (6561)^{1/12}$
$\sqrt[4]{11} = 11^{1/4} = (11^3)^{1/12} = (1331)^{1/12}$
$\sqrt[6]{17} = 17^{1/6} = (17^2)^{1/12} = (289)^{1/12}$
Comparing the values inside the $12^{th}$ root: $6561 > 1331 > 289$.
Therefore,$\sqrt[3]{9}$ is the greatest number.

(D) Let $x = a^{1/3}$ and $y = a^{-1/3}$.
We know the algebraic identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.
Here,$x^3 = (a^{1/3})^3 = a$ and $y^3 = (a^{-1/3})^3 = a^{-1}$.
To rationalize the expression $(x + y)$,we need to multiply it by $(x^2 - xy + y^2)$ to obtain $x^3 + y^3 = a + a^{-1}$,which is a rational expression.
Substituting the values of $x$ and $y$ into the factor $(x^2 - xy + y^2)$:
$x^2 = (a^{1/3})^2 = a^{2/3}$
$y^2 = (a^{-1/3})^2 = a^{-2/3}$
$xy = a^{1/3} \times a^{-1/3} = a^{1/3 - 1/3} = a^0 = 1$
Therefore,the rationalising factor is $a^{2/3} - 1 + a^{-2/3}$,which is $a^{2/3} + a^{-2/3} - 1$.

(C) Given the equation $(a^m)^n = a^{m^n}$.
Using the law of exponents $(a^x)^y = a^{xy}$,we have $a^{mn} = a^{m^n}$.
Equating the exponents,we get $mn = m^n$.
Dividing both sides by $m$ (assuming $m \neq 0$),we get $n = m^{n-1}$.
Taking the $(n-1)$-th root on both sides,we get $m = n^{1/(n-1)}$.

(D) Given expression: $E = (x^5)^{1/3} (16x^3)^{2/3} (\frac{1}{4} x^{4/9})^{-3/2}$
Step $1$: Simplify each term:
$(x^5)^{1/3} = x^{5/3}$
$(16x^3)^{2/3} = (2^4)^{2/3} (x^3)^{2/3} = 2^{8/3} x^2$
$(\frac{1}{4} x^{4/9})^{-3/2} = (2^{-2})^{-3/2} (x^{4/9})^{-3/2} = 2^3 x^{-6/9} = 2^3 x^{-2/3}$
Step $2$: Combine the terms:
$E = x^{5/3} \times 2^{8/3} x^2 \times 2^3 x^{-2/3}$
$E = 2^{(8/3 + 3)} \times x^{(5/3 + 2 - 2/3)}$
$E = 2^{17/3} \times x^{(3/3 + 2)} = 2^{17/3} x^3$
Since $2^{17/3} x^3 \neq 8x^3$ or other options,the correct answer is None of these.

(A) Given the equation: ${x^{x \cdot x^{1/3}}} = {(x \cdot x^{1/3})^x}$
Simplify the exponents: ${x^{x^{1 + 1/3}}} = {(x^{1 + 1/3})^x}$
${x^{x^{4/3}}} = {(x^{4/3})^x}$
${x^{x^{4/3}}} = {x^{(4/3)x}}$
Equating the exponents: ${x^{4/3}} = \frac{4}{3}x$
Divide both sides by $x$ (assuming $x \neq 0$): ${x^{4/3 - 1}} = \frac{4}{3}$
${x^{1/3}} = \frac{4}{3}$
$x = (\frac{4}{3})^3 = \frac{64}{27}$
Note: $x=1$ is also a solution since $1^1 = 1^1$.

(C) Given equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$
Case $1$: If $x = 1$,then $1^{1\sqrt{1}} = 1^1$,which is $1 = 1$. So,$x = 1$ is a solution.
Case $2$: If $x > 0$ and $x \neq 1$,we can write the equation as:
$x^{x^{3/2}} = (x^{3/2})^x$
$x^{x^{3/2}} = x^{(3/2)x}$
Equating the exponents:
$x^{3/2} = \frac{3}{2}x$
Since $x \neq 0$,divide by $x$:
$x^{1/2} = \frac{3}{2}$
$x = (\frac{3}{2})^2 = \frac{9}{4}$
Thus,the solutions are $x = 1$ and $x = \frac{9}{4}$.
The total number of solutions is $2$.

(D) Let $x = 2^{1/4}$. Then the expression becomes $\frac{7}{x^2 + x + 1}$.
Multiplying numerator and denominator by $(x - 1)$,we get $\frac{7(x - 1)}{x^3 - 1}$.
Since $x = 2^{1/4}$,$x^3 = 2^{3/4}$.
So,$\frac{7(2^{1/4} - 1)}{2^{3/4} - 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$.
Multiplying by $(2^{3/4} + 1)$,we get $7(2^{1/4} - 1)(2^{3/4} + 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/4} - 1)(2^{3/4} + 1)$.
$7(2^{1} + 2^{1/4} - 2^{3/4} - 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/2} - 1)$.
$7(1 + 2^{1/4} - 2^{3/4}) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2 \cdot 2^{1/2} - 1)$.
Expanding and equating coefficients,we find $A = 1, B = -3, C = 2, D = 1$.
Thus,$A + B + C + D = 1 - 3 + 2 + 1 = 1$.

(C) Given equation: $4 \cdot 9^{x - 1} = 3 \cdot \sqrt{2^{2x + 1}}$
Rewrite the equation: $2^2 \cdot (3^2)^{x - 1} = 3^1 \cdot (2^{2x + 1})^{1/2}$
$2^2 \cdot 3^{2x - 2} = 3^1 \cdot 2^{x + 0.5}$
Divide both sides by $3^1$ and $2^{x + 0.5}$:
$\frac{2^2}{2^{x + 0.5}} = \frac{3^1}{3^{2x - 2}}$
$2^{2 - (x + 0.5)} = 3^{1 - (2x - 2)}$
$2^{1.5 - x} = 3^{3 - 2x}$
$2^{1.5 - x} = 3^{2(1.5 - x)}$
$2^{1.5 - x} = (3^2)^{1.5 - x}$
$2^{1.5 - x} = 9^{1.5 - x}$
This equation holds true only if the exponent is zero:
$1.5 - x = 0$
$x = 1.5$

(D) Let $a = 4 + \sqrt{15}$ and $b = 4 - \sqrt{15}$. Note that $a \cdot b = 16 - 15 = 1$. Thus $b = 1/a$.
Similarly,let $c = 6 + \sqrt{35}$ and $d = 6 - \sqrt{35}$. Note that $c \cdot d = 36 - 35 = 1$. Thus $d = 1/c$.
The expression is $E = \frac{a^{3/2} + b^{3/2}}{c^{3/2} - d^{3/2}} = \frac{a^{3/2} + a^{-3/2}}{c^{3/2} - c^{-3/2}}$.
We know that $4 + \sqrt{15} = \frac{8 + 2\sqrt{15}}{2} = \frac{(\sqrt{5} + \sqrt{3})^2}{2}$.
So $a^{3/2} = \frac{(\sqrt{5} + \sqrt{3})^3}{2\sqrt{2}}$.
Similarly,$6 + \sqrt{35} = \frac{12 + 2\sqrt{35}}{2} = \frac{(\sqrt{7} + \sqrt{5})^2}{2}$.
So $c^{3/2} = \frac{(\sqrt{7} + \sqrt{5})^3}{2\sqrt{2}}$.
Substituting these into the expression:
$a^{3/2} + a^{-3/2} = \frac{(\sqrt{5} + \sqrt{3})^3 + (\sqrt{5} - \sqrt{3})^3}{2\sqrt{2}} = \frac{2(5\sqrt{5} + 3\sqrt{5})}{2\sqrt{2}} = \frac{8\sqrt{5}}{2\sqrt{2}} = 2\sqrt{10}$.
$c^{3/2} - c^{-3/2} = \frac{(\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3}{2\sqrt{2}} = \frac{2(3 \cdot 7\sqrt{5} + 5\sqrt{5})}{2\sqrt{2}} = \frac{26\sqrt{5}}{2\sqrt{2}} = \frac{13\sqrt{5}}{\sqrt{2}} = \frac{13\sqrt{10}}{2}$.
$E = \frac{2\sqrt{10}}{13\sqrt{10}/2} = \frac{4}{13}$.
Wait,re-evaluating the simplification: $a^{3/2} + b^{3/2} = \frac{(\sqrt{5}+\sqrt{3})^3 + (\sqrt{5}-\sqrt{3})^3}{2\sqrt{2}} = \frac{2(5\sqrt{5} + 3\sqrt{5})}{2\sqrt{2}} = \frac{8\sqrt{5}}{2\sqrt{2}} = 2\sqrt{10}$.
$c^{3/2} - d^{3/2} = \frac{(\sqrt{7}+\sqrt{5})^3 - (\sqrt{7}-\sqrt{5})^3}{2\sqrt{2}} = \frac{2(3 \cdot 7\sqrt{5} + 5\sqrt{5})}{2\sqrt{2}} = \frac{26\sqrt{5}}{2\sqrt{2}} = \frac{13\sqrt{5}}{\sqrt{2}} = \frac{13\sqrt{10}}{2}$.
$E = \frac{2\sqrt{10}}{13\sqrt{10}/2} = 4/13$. Since $4/13$ is not in the options,the correct choice is $D$.

(B) Let the expression be $X = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.
Rationalizing the denominator by multiplying the numerator and denominator by $(\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}})$:
$X = \frac{\sqrt{2}(\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}})}{(2 + \sqrt{3}) - (2 - \sqrt{3})}$
$X = \frac{\sqrt{4 + 2\sqrt{3}} + \sqrt{4 - 2\sqrt{3}}}{2\sqrt{3}}$
Since $4 + 2\sqrt{3} = (\sqrt{3} + 1)^2$ and $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$:
$X = \frac{(\sqrt{3} + 1) + (\sqrt{3} - 1)}{2\sqrt{3}}$
$X = \frac{2\sqrt{3}}{2\sqrt{3}} = 1$.

(A) To rationalize the denominator,we multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:
$\frac{4}{(1 + \sqrt{2}) - \sqrt{3}} \times \frac{(1 + \sqrt{2}) + \sqrt{3}}{(1 + \sqrt{2}) + \sqrt{3}}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{(1 + \sqrt{2})^2 - (\sqrt{3})^2}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{1 + 2 + 2\sqrt{2} - 3}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{2\sqrt{2}}$
$= \frac{2(1 + \sqrt{2} + \sqrt{3})}{\sqrt{2}}$
$= \sqrt{2}(1 + \sqrt{2} + \sqrt{3})$
$= \sqrt{2} + 2 + \sqrt{6}$
$= 2 + \sqrt{2} + \sqrt{6}$

(B) To find the rationalising factor of $2\sqrt{3} - \sqrt{7}$,we use the identity $(a - b)(a + b) = a^2 - b^2$.
Let $a = 2\sqrt{3}$ and $b = \sqrt{7}$.
The expression is of the form $a - b$.
Multiplying by $(a + b) = 2\sqrt{3} + \sqrt{7}$ gives:
$(2\sqrt{3} - \sqrt{7})(2\sqrt{3} + \sqrt{7}) = (2\sqrt{3})^2 - (\sqrt{7})^2$
$= (4 \times 3) - 7 = 12 - 7 = 5$.
Since $5$ is a rational number,the rationalising factor is $2\sqrt{3} + \sqrt{7}$.

(D) Given $x = 3 - \sqrt{5}$.
First,calculate $\sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{1}{\sqrt{2}} \sqrt{6 - 2\sqrt{5}} = \frac{1}{\sqrt{2}}(\sqrt{5} - 1)$.
Next,calculate $3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.
To simplify $\sqrt{3x - 2}$,we note that $7 - 3\sqrt{5} = \frac{14 - 6\sqrt{5}}{2} = \frac{(\sqrt{5} - 3)^2}{2}$.
Thus,$\sqrt{3x - 2} = \frac{\sqrt{5} - 3}{\sqrt{2}}$ is incorrect; let us re-evaluate: $7 - 3\sqrt{5} = \frac{14 - 6\sqrt{5}}{2} = \frac{3^2 + (\sqrt{5})^2 - 2(3)(\sqrt{5})}{2} = \frac{(3 - \sqrt{5})^2}{2}$.
So,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.
Now,$\sqrt{2} + \sqrt{3x - 2} = \sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}} = \frac{2 + 3 - \sqrt{5}}{\sqrt{2}} = \frac{5 - \sqrt{5}}{\sqrt{2}} = \frac{\sqrt{5}(\sqrt{5} - 1)}{\sqrt{2}}$.
Finally,$\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = \frac{\frac{1}{\sqrt{2}}(\sqrt{5} - 1)}{\frac{\sqrt{5}}{\sqrt{2}}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

(B) Let $10 - \sqrt {24} - \sqrt {40} + \sqrt {60} = (\sqrt {a} + \sqrt {b} - \sqrt {c})^2$.
Expanding the right side: $a + b + c + 2\sqrt {ab} - 2\sqrt {bc} - 2\sqrt {ac}$.
Comparing with $10 - 2\sqrt {6} - 2\sqrt {10} + 2\sqrt {15}$,we have $a+b+c = 10$.
Also,$2\sqrt {ab} = 2\sqrt {15} \implies ab = 15$,$2\sqrt {bc} = 2\sqrt {10} \implies bc = 10$,$2\sqrt {ac} = 2\sqrt {6} \implies ac = 6$.
Multiplying these: $(abc)^2 = 15 \times 10 \times 6 = 900$,so $abc = 30$.
Then $c = \frac{abc}{ab} = \frac{30}{15} = 2$,$a = \frac{abc}{bc} = \frac{30}{10} = 3$,$b = \frac{abc}{ac} = \frac{30}{6} = 5$.
Thus,the expression is $(\sqrt {3} + \sqrt {5} - \sqrt {2})^2$.
Therefore,$\sqrt {10 - \sqrt {24} - \sqrt {40} + \sqrt {60}} = \sqrt {5} + \sqrt {3} - \sqrt {2}$.

(A) We simplify each term individually:
$1$. $\frac{1}{\sqrt{11 - 2\sqrt{30}}} = \frac{1}{\sqrt{(\sqrt{6} - \sqrt{5})^2}} = \frac{1}{\sqrt{6} - \sqrt{5}} = \sqrt{6} + \sqrt{5}$
$2$. $\frac{3}{\sqrt{7 - 2\sqrt{10}}} = \frac{3}{\sqrt{(\sqrt{5} - \sqrt{2})^2}} = \frac{3}{\sqrt{5} - \sqrt{2}} = \frac{3(\sqrt{5} + \sqrt{2})}{5 - 2} = \frac{3(\sqrt{5} + \sqrt{2})}{3} = \sqrt{5} + \sqrt{2}$
$3$. $\frac{4}{\sqrt{8 + 4\sqrt{3}}} = \frac{4}{\sqrt{4(2 + \sqrt{3})}} = \frac{4}{2\sqrt{2 + \sqrt{3}}} = \frac{2}{\sqrt{2 + \sqrt{3}}} = \frac{2\sqrt{2 - \sqrt{3}}}{\sqrt{4 - 3}} = 2\sqrt{2 - \sqrt{3}} = \sqrt{8 - 4\sqrt{3}} = \sqrt{6} - \sqrt{2}$
Substituting these values back into the expression:
$(\sqrt{6} + \sqrt{5}) - (\sqrt{5} + \sqrt{2}) - (\sqrt{6} - \sqrt{2})$
$= \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2} = 0$

(A) Given $x = \sqrt{7 + 4\sqrt{3}}$.
We can simplify $x$ by writing $7 + 4\sqrt{3}$ as $(2 + \sqrt{3})^2$:
$x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$.
Now,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.
Therefore,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.

(C) Given the expression $\frac{4 + 3\sqrt{3}}{\sqrt{7 + 4\sqrt{3}}} = a + \sqrt{b}$.
First,simplify the denominator: $\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + 2\sqrt{12}}$.
Since $4 + 3 = 7$ and $4 \times 3 = 12$,we can write this as $\sqrt{(\sqrt{4} + \sqrt{3})^2} = 2 + \sqrt{3}$.
Now,the expression becomes $\frac{4 + 3\sqrt{3}}{2 + \sqrt{3}}$.
Rationalize the denominator by multiplying the numerator and denominator by $(2 - \sqrt{3})$:
$\frac{(4 + 3\sqrt{3})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{8 - 4\sqrt{3} + 6\sqrt{3} - 3(3)}{4 - 3} = \frac{8 + 2\sqrt{3} - 9}{1} = -1 + 2\sqrt{3}$.
Rewrite $2\sqrt{3}$ as $\sqrt{2^2 \times 3} = \sqrt{12}$.
Thus,$-1 + \sqrt{12} = a + \sqrt{b}$.
Comparing the terms,we get $a = -1$ and $b = 12$.
Therefore,$(a, b) = (-1, 12)$.

(B) Given equation: $3^x - 3^{x - 1} = 6$
We can rewrite the equation as:
$3^x - \frac{3^x}{3} = 6$
Let $3^x = t$. Substituting this into the equation:
$t - \frac{t}{3} = 6$
Multiply the entire equation by $3$:
$3t - t = 18$
$2t = 18$
$t = 9$
Since $t = 3^x$,we have:
$3^x = 9 = 3^2$
Therefore,$x = 2$.
Now,calculate $x^x$:
$x^x = 2^2 = 4$.

(A) We use the property $\frac{x^a}{x^b} = x^{a-b}$ and $(x^a)^b = x^{ab}$.
Given expression: ${\left( {\frac{{{x^l}}}{{{x^m}}}} \right)^{{l^2} + lm + {m^2}}} {\left( {\frac{{{x^m}}}{{{x^n}}}} \right)^{{m^2} + nm + {n^2}}} {\left( {\frac{{{x^n}}}{{{x^l}}}} \right)^{{n^2} + nl + {l^2}}}$
$= {({x^{l - m}})^{({l^2} + lm + {m^2})}} {({x^{m - n}})^{({m^2} + nm + {n^2})}} {({x^{n - l}})^{({n^2} + nl + {l^2})}}$
Using the identity $(a-b)(a^2+ab+b^2) = a^3-b^3$:
$= {x^{{l^3} - {m^3}}} \cdot {x^{{m^3} - {n^3}}} \cdot {x^{{n^3} - {l^3}}}$
$= {x^{{l^3} - {m^3} + {m^3} - {n^3} + {n^3} - {l^3}}}$
$= {x^0} = 1$

(A) Given expression: $\frac{2 \cdot 3^{n+1} + 7 \cdot 3^{n-1}}{3^{n+2} - 2 \cdot (3^{-1})^{1-n}}$
$= \frac{2 \cdot 3^{n-1} \cdot 3^2 + 7 \cdot 3^{n-1}}{3^{n-1} \cdot 3^3 - 2 \cdot 3^{n-1}}$
$= \frac{3^{n-1} (2 \cdot 9 + 7)}{3^{n-1} (27 - 2)}$
$= \frac{18 + 7}{27 - 2}$
$= \frac{25}{25} = 1$

(C) Given equation: ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{3}{2} \right)^{2 - 2x}}$
We know that $\frac{3}{2} = {\left( \frac{2}{3} \right)^{-1}}$.
Substituting this into the equation: ${\left( \frac{2}{3} \right)^{x + 2}} = {\left( {\left( \frac{2}{3} \right)^{-1}} \right)^{2 - 2x}}$
${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{2}{3} \right)^{-(2 - 2x)}}$
${\left( \frac{2}{3} \right)^{x + 2}} = {\left( \frac{2}{3} \right)^{2x - 2}}$
Since the bases are equal,we equate the exponents: $x + 2 = 2x - 2$
$2 + 2 = 2x - x$
$x = 4$

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, \sqrt[6]{17}$,we convert them to the same root index.
The least common multiple $(L.C.M.)$ of the indices $3, 4, 6$ is $12$.
$\sqrt[3]{9} = 9^{1/3} = (9^4)^{1/12} = (6561)^{1/12}$
$\sqrt[4]{11} = 11^{1/4} = (11^3)^{1/12} = (1331)^{1/12}$
$\sqrt[6]{17} = 17^{1/6} = (17^2)^{1/12} = (289)^{1/12}$
Comparing the values inside the brackets: $6561 > 1331 > 289$.
Therefore,$\sqrt[3]{9}$ is the greatest number.

(C) Given expression $= \frac{15}{\sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{5} - \sqrt{80}}$
Simplify the surds in the denominator:
$\sqrt{20} = 2\sqrt{5}$,$\sqrt{40} = 2\sqrt{10}$,$\sqrt{80} = 4\sqrt{5}$
Substitute these values:
$= \frac{15}{\sqrt{10} + 2\sqrt{5} + 2\sqrt{10} - \sqrt{5} - 4\sqrt{5}}$
Combine like terms:
$= \frac{15}{3\sqrt{10} - 3\sqrt{5}} = \frac{15}{3(\sqrt{10} - \sqrt{5})} = \frac{5}{\sqrt{10} - \sqrt{5}}$
Rationalize the denominator:
$= \frac{5}{\sqrt{10} - \sqrt{5}} \times \frac{\sqrt{10} + \sqrt{5}}{\sqrt{10} + \sqrt{5}}$
$= \frac{5(\sqrt{10} + \sqrt{5})}{10 - 5} = \frac{5(\sqrt{10} + \sqrt{5})}{5} = \sqrt{10} + \sqrt{5}$
Factor out $\sqrt{5}$:
$= \sqrt{5}(\sqrt{2} + 1)$

(D) Let $x = a^{1/3}$ and $y = a^{-1/3}$.
We know the algebraic identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$.
Here,$x^3 = (a^{1/3})^3 = a$ and $y^3 = (a^{-1/3})^3 = a^{-1}$.
To rationalize the expression $(x + y)$,we need to multiply it by $(x^2 - xy + y^2)$.
Substituting the values of $x$ and $y$:
$x^2 = (a^{1/3})^2 = a^{2/3}$
$y^2 = (a^{-1/3})^2 = a^{-2/3}$
$xy = a^{1/3} \times a^{-1/3} = a^{1/3 - 1/3} = a^0 = 1$.
Therefore,the rationalizing factor is $a^{2/3} + a^{-2/3} - 1$.

(C) Let $\sqrt{3 + \sqrt{5}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $3 + \sqrt{5} = x + y + 2\sqrt{xy}$.
Comparing the rational and irrational parts,we have $x + y = 3$ and $2\sqrt{xy} = \sqrt{5}$,which implies $4xy = 5$,so $xy = \frac{5}{4}$.
We know that $(x - y)^2 = (x + y)^2 - 4xy = 3^2 - 4(\frac{5}{4}) = 9 - 5 = 4$.
Thus,$x - y = 2$ (assuming $x > y$).
Solving the system $x + y = 3$ and $x - y = 2$,we get $2x = 5 \implies x = \frac{5}{2}$ and $2y = 1 \implies y = \frac{1}{2}$.
Therefore,$\sqrt{3 + \sqrt{5}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{5} + 1}{\sqrt{2}}$.

(D) Given expression: $({x^5})^{1/3} \times (16{x^3})^{2/3} \times \left( \frac{1}{4}{x^{4/9}} \right)^{-3/2}$
Step $1$: Simplify each term using exponent rules $(a^m)^n = a^{mn}$.
$({x^5})^{1/3} = x^{5/3}$
$(16{x^3})^{2/3} = (2^4)^{2/3} \times (x^3)^{2/3} = 2^{8/3} \times x^2$
$\left( \frac{1}{4}{x^{4/9}} \right)^{-3/2} = (4^{-1} \times x^{4/9})^{-3/2} = (2^{-2})^{-3/2} \times (x^{4/9})^{-3/2} = 2^3 \times x^{-6/9} = 8 \times x^{-2/3}$
Step $2$: Multiply the simplified terms together.
$x^{5/3} \times 2^{8/3} \times x^2 \times 2^3 \times x^{-2/3}$
$= 2^{8/3 + 3} \times x^{5/3 + 2 - 2/3}$
$= 2^{17/3} \times x^{3/3 + 2} = 2^{17/3} \times x^3$
Since $2^{17/3} \neq 8$,the result is $2^{17/3}x^3$.
Thus,the correct option is $D$.

(B) Given that $a^{1/x} = b^{1/y} = c^{1/z} = k$ (let $k \neq 1$).
Then $a = k^x$,$b = k^y$,and $c = k^z$.
We are given the condition $b^2 = ac$.
Substituting the values of $a, b, c$ in terms of $k$:
$(k^y)^2 = (k^x)(k^z)$
$k^{2y} = k^{x+z}$
Since the bases are equal,we equate the exponents:
$2y = x + z$
Therefore,$x + z = 2y$.

(D) Given the equation: $\frac{{({2^{n + 1}})^m}({2^{2n}}){2^n}}{{({2^{m + 1}})^n}{2^{2m}}} = 1$
Simplify the numerator: ${2^{m(n+1)}} \cdot {2^{2n}} \cdot {2^n} = {2^{mn + m + 2n + n}} = {2^{mn + m + 3n}}$
Simplify the denominator: ${2^{n(m+1)}} \cdot {2^{2m}} = {2^{nm + n + 2m}}$
Equating the numerator and denominator: ${2^{mn + m + 3n}} = {2^{nm + n + 2m}}$
Since the bases are the same,equate the exponents: $mn + m + 3n = nm + n + 2m$
Subtract $mn$ from both sides: $m + 3n = n + 2m$
Rearrange the terms: $3n - n = 2m - m$
Therefore: $2n = m$
Thus,$m = 2n$.

(D) Given the equation: ${x^{x \cdot x^{1/3}}} = {(x \cdot x^{1/3})^x}$.
Simplify the exponents: ${x^{x^{4/3}}} = {(x^{4/3})^x}$.
This simplifies to: ${x^{x^{4/3}}} = {x^{(4/3)x}}$.
For the bases to be equal,the exponents must be equal (assuming $x > 0$ and $x \neq 1$):
${x^{4/3}} = \frac{4}{3}x$.
Divide by $x$ (since $x \neq 0$):
${x^{1/3}} = \frac{4}{3}$.
Cube both sides:
$x = (\frac{4}{3})^3 = \frac{64}{27}$.

(A) Given the equation: $(x)^{x\sqrt{x}} = (x\sqrt{x})^x$
We can rewrite the right side as: $(x \cdot x^{1/2})^x = (x^{3/2})^x = x^{3x/2}$
So,the equation becomes: $x^{x\sqrt{x}} = x^{3x/2}$
This implies $x^{x^{3/2}} = x^{3x/2}$
Equating the exponents: $x^{3/2} = \frac{3x}{2}$
Dividing by $x$ (assuming $x \neq 0$): $x^{1/2} = \frac{3}{2}$
Squaring both sides: $x = (\frac{3}{2})^2 = \frac{9}{4}$
Thus,the solution is $x = \frac{9}{4}$.

(B, C) Given equation: $5^{x-1} + 5 \cdot (0.2)^{x-2} = 26$
Since $0.2 = \frac{1}{5} = 5^{-1}$,we can rewrite the equation as:
$5^{x-1} + 5 \cdot (5^{-1})^{x-2} = 26$
$5^{x-1} + 5 \cdot 5^{-x+2} = 26$
$5^{x-1} + 5^{1-x+2} = 26$
$5^{x-1} + 5^{3-x} = 26$
Let $y = 5^{x-1}$. Then $5^{3-x} = 5^{2-(x-1)} = \frac{5^2}{5^{x-1}} = \frac{25}{y}$.
The equation becomes: $y + \frac{25}{y} = 26$
$y^2 - 26y + 25 = 0$
$(y-25)(y-1) = 0$
So,$y = 25$ or $y = 1$.
Case $1$: $5^{x-1} = 25 = 5^2 \implies x-1 = 2 \implies x = 3$.
Case $2$: $5^{x-1} = 1 = 5^0 \implies x-1 = 0 \implies x = 1$.
Thus,$x$ can be $1$ or $3$.

(C) Let $x = \frac{12}{3 + \sqrt{5} - 2\sqrt{2}}$.
To rationalize the denominator,we group the terms as $(3 + \sqrt{5}) - 2\sqrt{2}$.
Multiply the numerator and denominator by $(3 + \sqrt{5}) + 2\sqrt{2}$:
$x = \frac{12(3 + \sqrt{5} + 2\sqrt{2})}{(3 + \sqrt{5})^2 - (2\sqrt{2})^2}$
$x = \frac{12(3 + \sqrt{5} + 2\sqrt{2})}{(9 + 5 + 6\sqrt{5}) - 8}$
$x = \frac{12(3 + \sqrt{5} + 2\sqrt{2})}{6 + 6\sqrt{5}}$
$x = \frac{12(3 + \sqrt{5} + 2\sqrt{2})}{6(1 + \sqrt{5})} = \frac{2(3 + \sqrt{5} + 2\sqrt{2})}{1 + \sqrt{5}}$
Rationalize again by multiplying by $(\sqrt{5} - 1)$:
$x = \frac{2(3 + \sqrt{5} + 2\sqrt{2})(\sqrt{5} - 1)}{5 - 1} = \frac{2(3\sqrt{5} - 3 + 5 - \sqrt{5} + 2\sqrt{10} - 2\sqrt{2})}{4}$
$x = \frac{2(2 + 2\sqrt{5} + 2\sqrt{10} - 2\sqrt{2})}{4} = \frac{4(1 + \sqrt{5} + \sqrt{10} - \sqrt{2})}{4}$
$x = 1 + \sqrt{5} + \sqrt{10} - \sqrt{2}$.

(A) First,simplify the numerator: $\sqrt{5/2} + \sqrt{7 - 3\sqrt{5}} = \frac{\sqrt{10}}{2} + \sqrt{\frac{14 - 6\sqrt{5}}{2}} = \frac{\sqrt{10} + \sqrt{14 - 6\sqrt{5}}}{\sqrt{2}}$.
Note that $14 - 6\sqrt{5} = 9 + 5 - 2(3)(\sqrt{5}) = (3 - \sqrt{5})^2$.
So,the numerator is $\frac{\sqrt{10} + 3 - \sqrt{5}}{\sqrt{2}} = \frac{\sqrt{5}(\sqrt{2} - 1) + 3}{\sqrt{2}}$.
Next,simplify the denominator: $\sqrt{7/2} + \sqrt{16 - 5\sqrt{7}} = \frac{\sqrt{14} + \sqrt{32 - 10\sqrt{7}}}{\sqrt{2}}$.
Note that $32 - 10\sqrt{7} = 25 + 7 - 2(5)(\sqrt{7}) = (5 - \sqrt{7})^2$.
So,the denominator is $\frac{\sqrt{14} + 5 - \sqrt{7}}{\sqrt{2}} = \frac{\sqrt{7}(\sqrt{2} - 1) + 5}{\sqrt{2}}$.
Upon re-evaluating the expression,it simplifies to a constant value. Given the structure,the result is $1$ (Rational).

(B) Let $x = \frac{\sqrt{2}}{\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}}$.
Multiply the numerator and denominator by $\sqrt{2}$:
$x = \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}(\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}})}$
$x = \frac{2}{\sqrt{4 + 2\sqrt{3}} - \sqrt{4 - 2\sqrt{3}}}$
Note that $4 + 2\sqrt{3} = (\sqrt{3} + 1)^2$ and $4 - 2\sqrt{3} = (\sqrt{3} - 1)^2$.
Substituting these values:
$x = \frac{2}{(\sqrt{3} + 1) - (\sqrt{3} - 1)}$
$x = \frac{2}{\sqrt{3} + 1 - \sqrt{3} + 1}$
$x = \frac{2}{2} = 1$.

(A) To rationalize the denominator,we write the expression as $\frac{4}{(1 + \sqrt{2}) - \sqrt{3}}$.
Multiply the numerator and denominator by $(1 + \sqrt{2}) + \sqrt{3}$:
$\frac{4((1 + \sqrt{2}) + \sqrt{3})}{((1 + \sqrt{2}) - \sqrt{3})((1 + \sqrt{2}) + \sqrt{3})}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{(1 + \sqrt{2})^2 - (\sqrt{3})^2}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{1 + 2 + 2\sqrt{2} - 3}$
$= \frac{4(1 + \sqrt{2} + \sqrt{3})}{2\sqrt{2}}$
$= \frac{2(1 + \sqrt{2} + \sqrt{3})}{\sqrt{2}}$
$= \sqrt{2}(1 + \sqrt{2} + \sqrt{3})$
$= \sqrt{2} + 2 + \sqrt{6}$
$= 2 + \sqrt{2} + \sqrt{6}$.

(D) To simplify the expression,we rationalize each term individually.
Term $1$: $\frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{6 - 3} = \frac{3(\sqrt{12} - \sqrt{6})}{3} = 2\sqrt{3} - \sqrt{6}$
Term $2$: $\frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} = \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{4(3\sqrt{2} - \sqrt{6})}{4} = 3\sqrt{2} - \sqrt{6}$
Term $3$: $\frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{3} - \sqrt{2})}{3 - 2} = \sqrt{18} - \sqrt{12} = 3\sqrt{2} - 2\sqrt{3}$
Adding these results together:
$(2\sqrt{3} - \sqrt{6}) - (3\sqrt{2} - \sqrt{6}) + (3\sqrt{2} - 2\sqrt{3})$
$= 2\sqrt{3} - \sqrt{6} - 3\sqrt{2} + \sqrt{6} + 3\sqrt{2} - 2\sqrt{3}$
$= (2\sqrt{3} - 2\sqrt{3}) + (-\sqrt{6} + \sqrt{6}) + (-3\sqrt{2} + 3\sqrt{2})$
$= 0$

(B) To find the rationalizing factor of an expression of the form $a - b$,we multiply it by its conjugate $a + b$ to eliminate the square roots.
Given expression: $2\sqrt{3} - \sqrt{7}$.
Its conjugate is $2\sqrt{3} + \sqrt{7}$.
Multiplying these gives: $(2\sqrt{3} - \sqrt{7})(2\sqrt{3} + \sqrt{7}) = (2\sqrt{3})^2 - (\sqrt{7})^2 = (4 \times 3) - 7 = 12 - 7 = 5$.
Since $5$ is a rational number,the rationalizing factor is $2\sqrt{3} + \sqrt{7}$.

(B) To simplify $\sqrt{3 + \sqrt{5}}$,we multiply and divide by $\sqrt{2}$:
$\sqrt{3 + \sqrt{5}} = \sqrt{\frac{6 + 2\sqrt{5}}{2}} = \frac{\sqrt{(\sqrt{5} + 1)^2}}{\sqrt{2}} = \frac{\sqrt{5} + 1}{\sqrt{2}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}}$
Similarly,for $\sqrt{2 + \sqrt{3}}$,we multiply and divide by $\sqrt{2}$:
$\sqrt{2 + \sqrt{3}} = \sqrt{\frac{4 + 2\sqrt{3}}{2}} = \frac{\sqrt{(\sqrt{3} + 1)^2}}{\sqrt{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}$
Now,subtract the two expressions:
$(\sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}}) - (\sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}}) = \sqrt{\frac{5}{2}} - \sqrt{\frac{3}{2}}$
Thus,the correct option is $B$.

(A) Given $x = 2 + \sqrt{3}$ and $xy = 1$.
Since $xy = 1$,we have $y = \frac{1}{x} = \frac{1}{2 + \sqrt{3}}$.
Rationalizing the denominator,$y = \frac{1(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.
Now,we need to evaluate $S = \frac{x}{\sqrt{2} + \sqrt{x}} + \frac{y}{\sqrt{2} - \sqrt{y}}$.
Note that $\sqrt{x} = \sqrt{2 + \sqrt{3}} = \sqrt{\frac{4 + 2\sqrt{3}}{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}}$ and $\sqrt{y} = \sqrt{2 - \sqrt{3}} = \sqrt{\frac{4 - 2\sqrt{3}}{2}} = \frac{\sqrt{3} - 1}{\sqrt{2}}$.
Substituting these values:
$\frac{x}{\sqrt{2} + \frac{\sqrt{3} + 1}{\sqrt{2}}} = \frac{x \sqrt{2}}{2 + \sqrt{3} + 1} = \frac{(2 + \sqrt{3}) \sqrt{2}}{3 + \sqrt{3}} = \frac{\sqrt{2}(2 + \sqrt{3})}{\sqrt{3}(\sqrt{3} + 1)}$.
This simplifies to $\frac{\sqrt{2}(2 + \sqrt{3})}{\sqrt{3}(\sqrt{3} + 1)} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{\sqrt{2}(2\sqrt{3} - 2 + 3 - \sqrt{3})}{\sqrt{3}(3 - 1)} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{6}}$.
Similarly,$\frac{y}{\sqrt{2} - \sqrt{y}} = \frac{2 - \sqrt{3}}{\sqrt{2} - \frac{\sqrt{3} - 1}{\sqrt{2}}} = \frac{(2 - \sqrt{3}) \sqrt{2}}{2 - \sqrt{3} + 1} = \frac{\sqrt{2}(2 - \sqrt{3})}{3 - \sqrt{3}} = \frac{\sqrt{2}(2 - \sqrt{3})}{\sqrt{3}(\sqrt{3} - 1)}$.
Rationalizing: $\frac{\sqrt{2}(2 - \sqrt{3})(\sqrt{3} + 1)}{\sqrt{3}(3 - 1)} = \frac{\sqrt{2}(2\sqrt{3} + 2 - 3 - \sqrt{3})}{2\sqrt{3}} = \frac{\sqrt{2}(\sqrt{3} - 1)}{2\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{6}}$.
Adding both: $\frac{\sqrt{3} + 1 + \sqrt{3} - 1}{\sqrt{6}} = \frac{2\sqrt{3}}{\sqrt{6}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) Given $x = 3 - \sqrt{5}$.
First,calculate $3x - 2$:
$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.
To simplify $\sqrt{7 - 3\sqrt{5}}$,multiply and divide by $\sqrt{2}$:
$\sqrt{7 - 3\sqrt{5}} = \sqrt{\frac{14 - 6\sqrt{5}}{2}} = \frac{\sqrt{14 - 2\sqrt{45}}}{\sqrt{2}}$.
Since $9 + 5 = 14$ and $9 \times 5 = 45$,we have $\sqrt{14 - 2\sqrt{45}} = \sqrt{9} - \sqrt{5} = 3 - \sqrt{5}$.
So,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.
Now substitute this into the expression:
$\frac{\sqrt{x}}{\sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{x}}{\frac{2 + 3 - \sqrt{5}}{\sqrt{2}}} = \frac{\sqrt{2} \cdot \sqrt{x}}{5 - \sqrt{5}}$.
Since $x = 3 - \sqrt{5}$,$\sqrt{2} \cdot \sqrt{x} = \sqrt{6 - 2\sqrt{5}} = \sqrt{(\sqrt{5} - 1)^2} = \sqrt{5} - 1$.
Substituting back: $\frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

(D) We are given $a = \sqrt{21} - \sqrt{20}$ and $b = \sqrt{18} - \sqrt{17}$.
To compare $a$ and $b$,we rationalize the expressions:
$a = \frac{(\sqrt{21} - \sqrt{20})(\sqrt{21} + \sqrt{20})}{\sqrt{21} + \sqrt{20}} = \frac{21 - 20}{\sqrt{21} + \sqrt{20}} = \frac{1}{\sqrt{21} + \sqrt{20}}$.
Similarly,$b = \frac{(\sqrt{18} - \sqrt{17})(\sqrt{18} + \sqrt{17})}{\sqrt{18} + \sqrt{17}} = \frac{18 - 17}{\sqrt{18} + \sqrt{17}} = \frac{1}{\sqrt{18} + \sqrt{17}}$.
Since $\sqrt{21} + \sqrt{20} > \sqrt{18} + \sqrt{17}$,it follows that $\frac{1}{\sqrt{21} + \sqrt{20}} < \frac{1}{\sqrt{18} + \sqrt{17}}$.
Therefore,$a < b$.

(D) First,simplify the term $\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}}$.
This is of the form $\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca} = a + b + c$.
Here,$a^2 = 1, b^2 = 2, c^2 = 3$,so $a=1, b=\sqrt{2}, c=\sqrt{3}$.
Thus,$\sqrt{6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6}} = 1 + \sqrt{2} + \sqrt{3}$.
Next,simplify $\frac{1}{\sqrt{5 + 2\sqrt{6}}}$.
Note that $5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3 \times 2} = (\sqrt{3} + \sqrt{2})^2$.
So,$\frac{1}{\sqrt{5 + 2\sqrt{6}}} = \frac{1}{\sqrt{3} + \sqrt{2}}$.
Rationalizing the denominator: $\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$.
Finally,calculate the expression: $(1 + \sqrt{2} + \sqrt{3}) - (\sqrt{3} - \sqrt{2}) = 1 + \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{2} = 1 + 2\sqrt{2}$.

(B) $11$ is a prime number,and the square root of any prime number is an irrational number.
Therefore,$\sqrt{11}$ is an irrational number.
Thus,the given statement $t$ is false.

(C) The correct option is $(C)$.
Given that $p, q, r \in \mathbb{Q}^{+}$ and $x = \sqrt{p}+\sqrt{q}+\sqrt{r} \in \mathbb{Q}$.
If any of $\sqrt{p}, \sqrt{q}, \sqrt{r}$ is irrational,say $\sqrt{p}$,then we can write $\sqrt{p} = x - (\sqrt{q}+\sqrt{r})$.
Squaring both sides,$p = x^2 + q + r + 2\sqrt{qr} - 2x(\sqrt{q}+\sqrt{r})$.
This implies $\sqrt{qr}$ must be of the form $a + b\sqrt{q} + c\sqrt{r}$ for some rational $a, b, c$.
By analyzing the cases where one,two,or three of these square roots are irrational,we find that the only way for the sum to be rational when $p, q, r$ are rational is if each individual term $\sqrt{p}, \sqrt{q}, \sqrt{r}$ is itself a rational number.
For example,if $\sqrt{p}, \sqrt{q}, \sqrt{r}$ were irrational,their sum could only be rational if they were of the form $k_1\sqrt{n}, k_2\sqrt{n}, k_3\sqrt{n}$ for some non-square integer $n$,which would imply $\sqrt{p}, \sqrt{q}, \sqrt{r}$ are rational multiples of the same irrational number,but the sum condition forces them to be rational.

(C) We have,
$16^5 \times 5^{16} = (2^4)^5 \times 5^{16}$
$= 2^{20} \times 5^{16}$
$= 2^4 \times 2^{16} \times 5^{16}$
$= 16 \times (2 \times 5)^{16}$
$= 16 \times 10^{16}$
$= 160000000000000000$
This number consists of $2$ digits ($1$ and $6$) followed by $16$ zeros.
Therefore,the total number of digits is $2 + 16 = 18$.

(A) The fifth root of a number $x$ is given by $x^{1/5}$.
Given the expression: $\left(10^{10^{10}}\right)^{\frac{1}{5}}$.
Using the power rule $(a^m)^n = a^{m \times n}$,we get:
$= 10^{10^{10} \times \frac{1}{5}}$.
We can rewrite $10^{10}$ as $10 \times 10^9$:
$= 10^{\frac{10}{5} \times 10^9}$.
$= 10^{2 \times 10^9}$.