Squares and Square Roots Questions in English

(C) First,rationalize $x$:
$x = \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{(\sqrt{3}+1)^{2}}{3-1} = \frac{3+1+2\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
Next,rationalize $y$:
$y = \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^{2}}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.
Now,calculate $x^{2}+y^{2}$:
$x^{2}+y^{2} = (2+\sqrt{3})^{2} + (2-\sqrt{3})^{2}$.
Using the identity $(a+b)^{2} + (a-b)^{2} = 2(a^{2}+b^{2})$:
$x^{2}+y^{2} = 2(2^{2} + (\sqrt{3})^{2}) = 2(4+3) = 2(7) = 14$.

(C) Given $x = 7 - 4\sqrt{3}$.
First,find the value of $\frac{1}{x}$ by rationalizing the denominator:
$\frac{1}{x} = \frac{1}{7 - 4\sqrt{3}} = \frac{1 \times (7 + 4\sqrt{3})}{(7 - 4\sqrt{3}) \times (7 + 4\sqrt{3})}$
Using the identity $(a - b)(a + b) = a^2 - b^2$:
$\frac{1}{x} = \frac{7 + 4\sqrt{3}}{7^2 - (4\sqrt{3})^2} = \frac{7 + 4\sqrt{3}}{49 - (16 \times 3)} = \frac{7 + 4\sqrt{3}}{49 - 48} = \frac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3}$
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (7 - 4\sqrt{3}) + (7 + 4\sqrt{3})$
$x + \frac{1}{x} = 7 + 7 - 4\sqrt{3} + 4\sqrt{3} = 14$.

(B) To find the square root of $\frac{\sqrt{2}-1}{\sqrt{2}+1}$,we first rationalize the denominator:
$\frac{\sqrt{2}-1}{\sqrt{2}+1} = \frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{(\sqrt{2}-1)^2}{2-1} = (\sqrt{2}-1)^2$.
Now,we need to find the square root of this expression:
$\sqrt{(\sqrt{2}-1)^2} = \sqrt{2}-1$.
Given $\sqrt{2} = 1.414$,we substitute this value:
$1.414 - 1 = 0.414$.

(C) To solve $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator,which is $(7-4 \sqrt{3})$.
$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}} = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 8(3)}{49 - 16(3)}$
$= \frac{35 - 24 - 6 \sqrt{3}}{49 - 48}$
$= \frac{11 - 6 \sqrt{3}}{1} = 11 - 6 \sqrt{3}$
Comparing $a+b \sqrt{3} = 11 - 6 \sqrt{3}$,we get $a = 11$ and $b = -6$.

(C) To solve the expression $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}-\frac{6}{\sqrt{8}-\sqrt{12}}$,we rationalize each term individually.
First term: $\frac{3 \sqrt{2}}{\sqrt{6}-\sqrt{3}} \times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}} = \frac{3 \sqrt{12}+3 \sqrt{6}}{6-3} = \frac{6 \sqrt{3}+3 \sqrt{6}}{3} = 2 \sqrt{3}+\sqrt{6}$.
Second term: $\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}} \times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} = \frac{4 \sqrt{18}+4 \sqrt{6}}{6-2} = \frac{12 \sqrt{2}+4 \sqrt{6}}{4} = 3 \sqrt{2}+\sqrt{6}$.
Third term: $\frac{6}{\sqrt{8}-\sqrt{12}} \times \frac{\sqrt{8}+\sqrt{12}}{\sqrt{8}+\sqrt{12}} = \frac{6(\sqrt{8}+\sqrt{12})}{8-12} = \frac{6(2 \sqrt{2}+2 \sqrt{3})}{-4} = -\frac{3(2 \sqrt{2}+2 \sqrt{3})}{2} = -3 \sqrt{2}-3 \sqrt{3}$.
Combining these: $(2 \sqrt{3}+\sqrt{6}) - (3 \sqrt{2}+\sqrt{6}) - (-3 \sqrt{2}-3 \sqrt{3})$.
$= 2 \sqrt{3}+\sqrt{6}-3 \sqrt{2}-\sqrt{6}+3 \sqrt{2}+3 \sqrt{3}$.
$= 2 \sqrt{3}+3 \sqrt{3} = 5 \sqrt{3}$.

(B) Given expression: $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$
Rationalizing the denominators:
$\frac{1}{2+\sqrt{2}} = \frac{1}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2-\sqrt{2}}{4-2} = \frac{2-\sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$
$\frac{1}{\sqrt{2}-2} = \frac{1}{\sqrt{2}-2} \times \frac{\sqrt{2}+2}{\sqrt{2}+2} = \frac{\sqrt{2}+2}{2-4} = \frac{\sqrt{2}+2}{-2} = -\frac{\sqrt{2}}{2} - 1$
Substituting these back into the expression:
$= 2 + \sqrt{2} + (1 - \frac{\sqrt{2}}{2}) + (-1 - \frac{\sqrt{2}}{2})$
$= 2 + \sqrt{2} + 1 - \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{2}}{2}$
$= 2 + (1 - 1) + (\sqrt{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2})$
$= 2 + 0 + (\sqrt{2} - \sqrt{2}) = 2$

(C) To solve $\frac{1}{\sqrt{5}-\sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{5}+\sqrt{3})$.
$\frac{1}{\sqrt{5}-\sqrt{3}} = \frac{1}{(\sqrt{5}-\sqrt{3})} \times \frac{(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2}$
$= \frac{\sqrt{5}+\sqrt{3}}{5-3} = \frac{\sqrt{5}+\sqrt{3}}{2}$
Substitute the given values $\sqrt{5} = 2.2361$ and $\sqrt{3} = 1.7321$:
$= \frac{2.2361 + 1.7321}{2} = \frac{3.9682}{2} = 1.9841$

(A) We can write the given number as $0.000326 = 326 \times 10^{-6}$.
To make this a perfect square,we look for the nearest perfect square less than $326$.
The nearest perfect square less than $326$ is $324$,which is the square of $18$ $(18^2 = 324)$.
Thus,$324 \times 10^{-6} = 0.000324$ is a perfect square.
The number to be subtracted is $0.000326 - 0.000324 = 0.000002$.

(B) To find the smallest number by which $5808$ should be multiplied to make it a perfect square,we first find its prime factorization.
$5808 = 2 \times 2904 = 2 \times 2 \times 1452 = 2 \times 2 \times 2 \times 726 = 2 \times 2 \times 2 \times 2 \times 363 = 2 \times 2 \times 2 \times 2 \times 3 \times 121 = 2^4 \times 3 \times 11^2$.
In a perfect square,every prime factor must have an even exponent.
Here,the prime factor $3$ has an exponent of $1$,which is odd.
To make the exponent of $3$ even (i.e.,$2$),we must multiply the number by $3$.
Therefore,the smallest number to be multiplied is $3$.

(B) Given expression: $\frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}} + \sqrt{125}$
Simplify each term:
$1$. $\frac{10}{\sqrt{5}} = \frac{10 \times \sqrt{5}}{5} = 2\sqrt{5}$
$2$. $\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$
Substitute these into the expression:
$\frac{\sqrt{5}}{2} - 2\sqrt{5} + 5\sqrt{5} = \frac{\sqrt{5}}{2} + 3\sqrt{5} = \sqrt{5} \left( \frac{1}{2} + 3 \right) = \frac{7}{2} \sqrt{5}$
Given $\sqrt{5} = 2.236$,substitute the value:
$\frac{7}{2} \times 2.236 = 7 \times 1.118 = 7.826$

(C) Given expression is $\frac{1+\sqrt{0.01}}{1-\sqrt{0.1}}$.
Since $\sqrt{0.01} = 0.1$,the expression becomes $\frac{1+0.1}{1-\sqrt{0.1}} = \frac{1.1}{1-\sqrt{0.1}}$.
To rationalize the denominator,multiply the numerator and denominator by $(1+\sqrt{0.1})$:
$\frac{1.1(1+\sqrt{0.1})}{(1-\sqrt{0.1})(1+\sqrt{0.1})} = \frac{1.1(1+\sqrt{0.1})}{1-0.1} = \frac{1.1(1+\sqrt{0.1})}{0.9}$.
Using the approximation $\sqrt{0.1} \approx 0.3162$:
$= \frac{1.1(1+0.3162)}{0.9} = \frac{1.1(1.3162)}{0.9} = \frac{1.44782}{0.9} \approx 1.6086$.
Rounding to one decimal place,the value is approximately $1.6$.

(A) Given the repeating decimal $0 . \overline{4}$.
We can express this as a fraction: $0 . \overline{4} = \frac{4}{9}$.
Now,we need to find the square root of this value: $\sqrt{0 . \overline{4}} = \sqrt{\frac{4}{9}}$.
Calculating the square root: $\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$.
Converting the fraction $\frac{2}{3}$ back to a decimal: $\frac{2}{3} = 0.666 \dots = 0 . \overline{6}$.
Therefore,the correct option is $A$.

(C) The given expression is $\sqrt{4a^{2} - 4a + 1} + 3a$.
First,simplify the term inside the square root: $4a^{2} - 4a + 1 = (2a - 1)^{2}$.
Thus,the expression becomes $\sqrt{(2a - 1)^{2}} + 3a = |2a - 1| + 3a$.
Since $a = 0.1039$,$2a = 0.2078$,which is less than $1$. Therefore,$|2a - 1| = -(2a - 1) = 1 - 2a$.
Substituting this back into the expression: $(1 - 2a) + 3a = a + 1$.
Now,substitute the value of $a$: $0.1039 + 1 = 1.1039$.

(C) Using the algebraic identity $(a-b)^2 = a^2 + b^2 - 2ab$,where $a = \sqrt{3}$ and $b = \frac{1}{\sqrt{3}}$:
$\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)^{2} = (\sqrt{3})^2 + \left(\frac{1}{\sqrt{3}}\right)^2 - 2 \times \sqrt{3} \times \frac{1}{\sqrt{3}}$
$= 3 + \frac{1}{3} - 2$
$= 1 + \frac{1}{3}$
$= \frac{4}{3}$

(B) We use the algebraic identity $(a+b)(a-b) = a^2 - b^2$.
Here,$a = 7$ and $b = 3 \sqrt{5}$.
So,$(7+3 \sqrt{5})(7-3 \sqrt{5}) = 7^2 - (3 \sqrt{5})^2$.
Calculating the squares: $7^2 = 49$ and $(3 \sqrt{5})^2 = 9 \times 5 = 45$.
Thus,$49 - 45 = 4$.
The square root of the result is $\sqrt{4} = 2$.

(B) We are given the expression: $\sqrt{\frac{(0.03)^{2}+(0.21)^{2}+(0.065)^{2}}{(0.003)^{2}+(0.021)^{2}+(0.0065)^{2}}}$
Notice that each term in the numerator is $10$ times the corresponding term in the denominator: $0.03 = 10 \times 0.003$,$0.21 = 10 \times 0.021$,and $0.065 = 10 \times 0.0065$.
Substituting these into the numerator: $(0.03)^{2} = (10 \times 0.003)^{2} = 10^{2} \times (0.003)^{2}$,$(0.21)^{2} = 10^{2} \times (0.021)^{2}$,and $(0.065)^{2} = 10^{2} \times (0.0065)^{2}$.
Factoring out $10^{2}$ from the numerator: $\sqrt{\frac{10^{2} \left[(0.003)^{2} + (0.021)^{2} + (0.0065)^{2}\right]}{(0.003)^{2} + (0.021)^{2} + (0.0065)^{2}}}$
The terms in the brackets cancel out: $\sqrt{10^{2}} = 10$.

(D) First,simplify the given expression: $3 \sqrt{5} + \sqrt{125} = 3 \sqrt{5} + \sqrt{25 \times 5} = 3 \sqrt{5} + 5 \sqrt{5} = 8 \sqrt{5}$.
Given that $8 \sqrt{5} = 17.88$,we can find the value of $\sqrt{5}$ as $\sqrt{5} = \frac{17.88}{8} = 2.235$.
Now,simplify the expression to be evaluated: $\sqrt{80} + 6 \sqrt{5} = \sqrt{16 \times 5} + 6 \sqrt{5} = 4 \sqrt{5} + 6 \sqrt{5} = 10 \sqrt{5}$.
Substitute the value of $\sqrt{5}$: $10 \times 2.235 = 22.35$.

(A) Given the equation: $\sqrt{1+\frac{55}{729}}=1+\frac{x}{27}$
First,simplify the expression inside the square root: $1+\frac{55}{729} = \frac{729+55}{729} = \frac{784}{729}$
Now,take the square root of the fraction: $\sqrt{\frac{784}{729}} = \frac{\sqrt{784}}{\sqrt{729}} = \frac{28}{27}$
Substitute this back into the equation: $\frac{28}{27} = 1+\frac{x}{27}$
Subtract $1$ from both sides: $\frac{28}{27} - 1 = \frac{x}{27}$
$\frac{28-27}{27} = \frac{x}{27}$
$\frac{1}{27} = \frac{x}{27}$
Therefore,$x = 1$.

(C) To solve $\sqrt{\frac{48.4}{0.289}}$,first remove the decimals by multiplying the numerator and denominator by $1000$:
$\sqrt{\frac{48.4 \times 1000}{0.289 \times 1000}} = \sqrt{\frac{48400}{289}}$
Since $\sqrt{48400} = 220$ and $\sqrt{289} = 17$,the expression becomes $\frac{220}{17}$.
Dividing $220$ by $17$ gives $12$ with a remainder of $16$.
Therefore,$\frac{220}{17} = 12 \frac{16}{17}$.

(A) Given the equation: $\sqrt{x} + \sqrt{441} = 0.02$
We know that $\sqrt{441} = 21$.
Substituting this value into the equation: $\sqrt{x} + 21 = 0.02$
$\sqrt{x} = 0.02 - 21$
$\sqrt{x} = -20.98$
Since the square root of a real number cannot be negative,there is no real value for $x$ that satisfies this equation. However,if the equation was intended to be $\sqrt{x} \times \sqrt{441} = 0.02$:
$\sqrt{x} \times 21 = 0.02$
$\sqrt{x} = \frac{0.02}{21} \approx 0.000952$
$x = (0.000952)^2 \approx 0.000000906$.
Given the provided options,it appears there is a typo in the question. If the equation was $\sqrt{x} = 0.02 \times \sqrt{441}$:
$\sqrt{x} = 0.02 \times 21 = 0.42$
$x = (0.42)^2 = 0.1764$.

(B) Given equation: $28 \sqrt{?} + 1426 = \frac{3}{4} \times 2872$
First,calculate $\frac{3}{4} \times 2872 = 3 \times 718 = 2154$.
Now,substitute this back into the equation: $28 \sqrt{?} + 1426 = 2154$.
Subtract $1426$ from both sides: $28 \sqrt{?} = 2154 - 1426 = 728$.
Divide by $28$: $\sqrt{?} = \frac{728}{28} = 26$.
Square both sides to find the value of $?$: $? = (26)^2 = 676$.

(A) Let the required number be $x$.
According to the problem,$\frac{x}{\sqrt{0.25}} = 25$.
We know that $\sqrt{0.25} = 0.5$.
Substituting this value into the equation,we get $\frac{x}{0.5} = 25$.
Therefore,$x = 25 \times 0.5 = 12.5$.

(D) Given equation: $\frac{52}{x} = \sqrt{\frac{169}{289}}$
First,calculate the square root of the fraction on the right side:
$\sqrt{\frac{169}{289}} = \frac{\sqrt{169}}{\sqrt{289}} = \frac{13}{17}$
Now,substitute this back into the equation:
$\frac{52}{x} = \frac{13}{17}$
Cross-multiply to solve for $x$:
$13x = 52 \times 17$
$x = \frac{52 \times 17}{13}$
Since $52 \div 13 = 4$,we get:
$x = 4 \times 17 = 68$

(B) To find the square root of $1.5625$,we can write it as a fraction:
$1.5625 = \frac{15625}{10000}$
Now,take the square root of the numerator and the denominator:
$\sqrt{\frac{15625}{10000}} = \frac{\sqrt{15625}}{\sqrt{10000}}$
Since $125^2 = 15625$ and $100^2 = 10000$,we have:
$\frac{125}{100} = 1.25$
Therefore,$\sqrt{1.5625} = 1.25$.

(B) Step $1$: Calculate the square of $1.5$.
$1.5^{2} = 1.5 \times 1.5 = 2.25$.
Step $2$: Calculate the square root of $0.0225$.
$\sqrt{0.0225} = \sqrt{\frac{225}{10000}} = \frac{15}{100} = 0.15$.
Step $3$: Multiply the results from Step $1$ and Step $2$.
$2.25 \times 0.15 = 0.3375$.

(C) To find the square root of $0.000441$,we can express the decimal as a fraction or in scientific notation.
$0.000441 = \frac{441}{1000000} = \frac{441}{10^6}$.
Taking the square root of both sides:
$\sqrt{0.000441} = \sqrt{\frac{441}{10^6}} = \frac{\sqrt{441}}{\sqrt{10^6}}$.
Since $\sqrt{441} = 21$ and $\sqrt{10^6} = 10^3 = 1000$,we have:
$\frac{21}{1000} = 0.021$.

(C) To find the unit's digit of the square root of $15876$,we observe the unit's digit of the number itself,which is $6$.
If a perfect square ends in $6$,its square root must end in either $4$ or $6$ (since $4^2 = 16$ and $6^2 = 36$).
We can estimate the square root: $100^2 = 10000$ and $200^2 = 40000$.
Since $15876$ is between $100^2$ and $200^2$,the square root is between $100$ and $200$.
Checking $120^2 = 14400$ and $130^2 = 16900$.
Since $15876$ is between $14400$ and $16900$,the square root is between $120$ and $130$.
The possible values are $124$ or $126$.
Calculating $126^2 = 15876$.
Therefore,the unit's digit is $6$.

(B) Given the operation $x * y = x + y + \sqrt{xy}$.
To find the value of $6 * 24$,substitute $x = 6$ and $y = 24$ into the given expression:
$6 * 24 = 6 + 24 + \sqrt{6 \times 24}$
$= 30 + \sqrt{144}$
$= 30 + 12$
$= 42$

(A) First,evaluate the square roots in the expression:
$\sqrt{625} = 25$
$\sqrt{25} = 5$
$\sqrt{196} = 14$
Substitute these values into the expression:
$\left(\frac{25}{11} \times \frac{14}{5} \times \frac{11}{14}\right)$
Now,cancel the common terms in the numerator and denominator:
$= \left(\frac{25}{5} \times \frac{14}{14} \times \frac{11}{11}\right)$
$= 5 \times 1 \times 1 = 5$

(C) To evaluate the expression $\sqrt{41-\sqrt{21+\sqrt{19-\sqrt{9}}}}$,we solve from the innermost square root outwards.
Step $1$: Evaluate $\sqrt{9} = 3$.
Step $2$: Substitute this into the next term: $\sqrt{19-3} = \sqrt{16} = 4$.
Step $3$: Substitute this into the next term: $\sqrt{21+4} = \sqrt{25} = 5$.
Step $4$: Finally,evaluate the outermost term: $\sqrt{41-5} = \sqrt{36} = 6$.
Thus,the value of the expression is $6$.

(A) To find the value of the expression,we evaluate the nested square roots from the innermost to the outermost:
Step $1$: Evaluate $\sqrt{225} = 15$.
Step $2$: Substitute this into the expression: $\sqrt{154 + 15} = \sqrt{169} = 13$.
Step $3$: Substitute this into the expression: $\sqrt{108 + 13} = \sqrt{121} = 11$.
Step $4$: Substitute this into the expression: $\sqrt{25 + 11} = \sqrt{36} = 6$.
Step $5$: Finally,evaluate: $\sqrt{10 + 6} = \sqrt{16} = 4$.
Thus,the value is $4$.

(C) To find the square root of $(272^{2}-128^{2})$,we use the algebraic identity $a^{2}-b^{2}=(a+b)(a-b)$.
Here,$a=272$ and $b=128$.
Substituting these values,we get:
$\sqrt{272^{2}-128^{2}} = \sqrt{(272+128)(272-128)}$
$= \sqrt{(400)(144)}$
$= \sqrt{400} \times \sqrt{144}$
$= 20 \times 12$
$= 240$

(B) To find the square root of $0.00004761$,we can express the number as a fraction or in scientific notation.
$0.00004761 = \frac{4761}{100000000}$.
Taking the square root of both the numerator and the denominator:
$\sqrt{0.00004761} = \sqrt{\frac{4761}{10^8}} = \frac{\sqrt{4761}}{\sqrt{10^8}}$.
We know that $69^2 = 4761$,so $\sqrt{4761} = 69$.
Also,$\sqrt{10^8} = 10^4 = 10000$.
Therefore,$\sqrt{0.00004761} = \frac{69}{10000} = 0.0069$.

(D) To find the value of the expression $\sqrt{0.01} + \sqrt{0.81} + \sqrt{1.21} + \sqrt{0.0009}$,we calculate each square root individually:
$1$. $\sqrt{0.01} = \sqrt{\frac{1}{100}} = 0.1$
$2$. $\sqrt{0.81} = \sqrt{\frac{81}{100}} = 0.9$
$3$. $\sqrt{1.21} = \sqrt{\frac{121}{100}} = 1.1$
$4$. $\sqrt{0.0009} = \sqrt{\frac{9}{10000}} = 0.03$
Adding these values together: $0.1 + 0.9 + 1.1 + 0.03 = 2.13$.

(B) To find the square root of $1.5625$,we can write it as a fraction:
$\sqrt{1.5625} = \sqrt{\frac{15625}{10000}}$
Since $\sqrt{15625} = 125$ and $\sqrt{10000} = 100$,we have:
$\frac{125}{100} = 1.25$
Therefore,$\sqrt{1.5625} = 1.25$.

(A) Let the number be $x$.
According to the problem,the equation is:
$\frac{3}{5} \times x^{2} = 126.15$
Multiply both sides by $\frac{5}{3}$ to isolate $x^{2}$:
$x^{2} = \frac{126.15 \times 5}{3}$
$x^{2} = 42.05 \times 5$
$x^{2} = 210.25$
Now,take the square root of both sides to find $x$:
$x = \sqrt{210.25}$
$x = 14.5$

(C) To solve $\sqrt{\frac{96.8}{0.578}}$,first remove the decimals by multiplying the numerator and denominator by $1000$:
$\sqrt{\frac{96.8 \times 1000}{0.578 \times 1000}} = \sqrt{\frac{96800}{578}}$
Divide both the numerator and denominator by $2$:
$\sqrt{\frac{48400}{289}}$
Calculate the square root of the numerator and denominator separately:
$\sqrt{48400} = 220$ and $\sqrt{289} = 17$
So,the fraction is $\frac{220}{17}$.
Convert the improper fraction to a mixed number:
$220 \div 17 = 12$ with a remainder of $16$.
Thus,the result is $12 \frac{16}{17}$.

(B) To find the value of the expression $\frac{\sqrt{80}-\sqrt{112}}{\sqrt{45}-\sqrt{63}}$,we simplify the square roots first:
$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$
$\sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7}$
$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$
$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$
Substituting these values into the expression:
$\frac{4\sqrt{5}-4\sqrt{7}}{3\sqrt{5}-3\sqrt{7}} = \frac{4(\sqrt{5}-\sqrt{7})}{3(\sqrt{5}-\sqrt{7})}$
Canceling the common term $(\sqrt{5}-\sqrt{7})$ from the numerator and denominator:
$= \frac{4}{3} = 1 \frac{1}{3}$

(A) To find the smallest number to be added to $680621$ to make it a perfect square,we first find the square root of $680621$ using the long division method.
Step $1$: Group the digits in pairs from right to left: $68, 06, 21$.
Step $2$: Find the largest square less than or equal to $68$,which is $8^2 = 64$. So,the first digit is $8$.
Step $3$: Subtract $64$ from $68$ to get $4$,and bring down the next pair $06$ to get $406$.
Step $4$: Double the quotient $8$ to get $16$. Find a digit $x$ such that $16x \times x \leq 406$. For $x = 2$,$162 \times 2 = 324$. For $x = 3$,$163 \times 3 = 489$ (too large).
Step $5$: Subtract $324$ from $406$ to get $82$,and bring down the next pair $21$ to get $8221$.
Step $6$: The current quotient is $82$. Double it to get $164$. Find a digit $y$ such that $164y \times y \leq 8221$. For $y = 4$,$1644 \times 4 = 6576$. For $y = 5$,$1645 \times 5 = 8225$.
Since $8225 > 8221$,the next perfect square is $825^2 = 680625$.
Step $7$: The number to be added is $680625 - 680621 = 4$.

(A) To simplify the expression,we rationalize each term individually:
Step $1$: Rationalize $\frac{2+\sqrt{3}}{2-\sqrt{3}}$:
$\frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{(2+\sqrt{3})^2}{4-3} = 4+3+4\sqrt{3} = 7+4\sqrt{3}$
Step $2$: Rationalize $\frac{2-\sqrt{3}}{2+\sqrt{3}}$:
$\frac{2-\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{(2-\sqrt{3})^2}{4-3} = 4+3-4\sqrt{3} = 7-4\sqrt{3}$
Step $3$: Rationalize $\frac{\sqrt{3}-1}{\sqrt{3}+1}$:
$\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$
Step $4$: Add the results:
$(7+4\sqrt{3}) + (7-4\sqrt{3}) + (2-\sqrt{3})$
$= 7 + 7 + 2 + 4\sqrt{3} - 4\sqrt{3} - \sqrt{3}$
$= 16 - \sqrt{3}$

(B) To find the least number to be subtracted,we express the number as $0.000365 = 365 \times 10^{-6}$.
We look for the nearest perfect square less than $365$. The perfect square nearest to $365$ is $361$,which is $19^2$.
So,$361 \times 10^{-6} = 0.000361$ is a perfect square.
The number to be subtracted is $0.000365 - 0.000361 = 0.000004$.

(A) To find the value of $\frac{1}{\sqrt{5}}$,we rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{5}$.
$\frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.
Given that $\sqrt{5} = 2.236$,we substitute this value into the expression:
$\frac{2.236}{5} = 0.4472$.
Rounding to three decimal places,we get $0.447$.

(D) First,calculate the square roots in the expression:
$\sqrt{0.04} = 0.2$
$\sqrt{0.1} \approx 0.3162$
Substitute these values into the expression:
$\frac{1 + 0.2}{1 - 0.3162} = \frac{1.2}{0.6838}$
Performing the division:
$\frac{1.2}{0.6838} \approx 1.754$
Rounding to the nearest given option,the value is approximately $1.7$.

(B) Let the expression be $E = \frac{(0.75)^{3}}{1-0.75} + [0.75 + 0.75^{2} + 1]$.
First,simplify the fraction: $\frac{(0.75)^{3}}{0.25} = \frac{(0.75)^{3}}{(0.75/3)} = 3 \times (0.75)^{2}$.
Now,substitute this back into the expression: $E = 3 \times (0.75)^{2} + 0.75^{2} + 0.75 + 1$.
Combine the terms: $E = 4 \times (0.75)^{2} + 0.75 + 1$.
Calculate the values: $4 \times (0.5625) + 0.75 + 1 = 2.25 + 0.75 + 1 = 4$.
The square root of the expression is $\sqrt{4} = 2$.

(B) First,simplify each radical term:
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
$2\sqrt{32} = 2\sqrt{16 \times 2} = 2 \times 4\sqrt{2} = 8\sqrt{2}$
$3\sqrt{128} = 3\sqrt{64 \times 2} = 3 \times 8\sqrt{2} = 24\sqrt{2}$
$4\sqrt{50} = 4\sqrt{25 \times 2} = 4 \times 5\sqrt{2} = 20\sqrt{2}$
Now,substitute these into the expression:
$2\sqrt{2} + 8\sqrt{2} - 24\sqrt{2} + 20\sqrt{2}$
$= (2 + 8 - 24 + 20)\sqrt{2}$
$= 6\sqrt{2}$
Given $\sqrt{2} = 1.414$,we have:
$6 \times 1.414 = 8.484$

(A) To find the least perfect square divisible by $21, 36$,and $66$,we first find their Least Common Multiple $(LCM)$.
Prime factorization of the numbers:
$21 = 3 \times 7$
$36 = 2^2 \times 3^2$
$66 = 2 \times 3 \times 11$
$LCM$ $= 2^2 \times 3^2 \times 7 \times 11 = 4 \times 9 \times 77 = 2772$.
For a number to be a perfect square,the exponent of each prime factor in its prime factorization must be even.
In the $LCM$ $(2^2 \times 3^2 \times 7^1 \times 11^1)$,the prime factors $7$ and $11$ have odd exponents.
To make it a perfect square,we must multiply the $LCM$ by $7 \times 11 = 77$.
Required perfect square $= 2772 \times 77 = 213444$.

(D) To find the value of $\sqrt{\frac{0.16}{0.4}}$,first simplify the fraction inside the square root.
$\frac{0.16}{0.4} = \frac{16}{40} = \frac{4}{10} = 0.4$.
Now,calculate the square root of $0.4$.
$\sqrt{0.4} \approx 0.63245$.
Comparing this with the given options,none of the values $0.02$,$0.2$,or $0.63$ are equal to $\sqrt{0.4}$.
Therefore,the correct option is $D$.

(B) To solve the expression $\sqrt{\frac{0.081 \times 0.484}{0.0064 \times 6.25}}$,we first adjust the decimal places to simplify the calculation.
$\sqrt{\frac{0.081 \times 0.484}{0.0064 \times 6.25}} = \sqrt{\frac{81 \times 10^{-3} \times 484 \times 10^{-3}}{64 \times 10^{-4} \times 625 \times 10^{-2}}}$
$= \sqrt{\frac{81 \times 484 \times 10^{-6}}{64 \times 625 \times 10^{-6}}}$
$= \sqrt{\frac{81 \times 484}{64 \times 625}}$
$= \frac{\sqrt{81} \times \sqrt{484}}{\sqrt{64} \times \sqrt{625}}$
$= \frac{9 \times 22}{8 \times 25}$
$= \frac{198}{200} = 0.99$

(B) Given the equation: $\frac{0.13}{p^{2}} = 13$
Rearranging the terms to solve for $p^{2}$,we get:
$p^{2} = \frac{0.13}{13}$
Performing the division:
$p^{2} = 0.01$
Taking the square root of both sides:
$p = \sqrt{0.01}$
$p = 0.1$

(A) Let the value in place of the question mark be $x$.
Then the equation becomes $\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x}$.
By cross-multiplying,we get $x^2 = \sqrt{128} \times \sqrt{162}$.
$x^2 = \sqrt{128 \times 162}$.
$x^2 = \sqrt{(64 \times 2) \times (81 \times 2)}$.
$x^2 = \sqrt{64 \times 81 \times 4}$.
$x^2 = 8 \times 9 \times 2$.
$x^2 = 144$.
$x = \sqrt{144} = 12$.