Squares and Square Roots Questions in English

(D) Given the equation: $\left(\frac{x}{15}\right)\left(\frac{x}{135}\right) = 1$
Multiply the numerators and denominators: $\frac{x^2}{15 \times 135} = 1$
$x^2 = 15 \times 135$
$x^2 = 15 \times (15 \times 9)$
$x^2 = 15^2 \times 3^2$
Taking the square root on both sides: $x = \sqrt{15^2 \times 3^2}$
$x = 15 \times 3 = 45$
Therefore,the value of $x$ is $45$.

(D) We need to evaluate $\sqrt{1.3} + \sqrt{1300} + \sqrt{0.013}$.
First,express each term in terms of $\sqrt{13}$ or $\sqrt{130}$:
$\sqrt{1.3} = \sqrt{\frac{130}{100}} = \frac{\sqrt{130}}{10} = 0.1 \times 11.4 = 1.14$
$\sqrt{1300} = \sqrt{100 \times 13} = 10 \times \sqrt{13} = 10 \times 3.605 = 36.05$
$\sqrt{0.013} = \sqrt{\frac{130}{10000}} = \frac{\sqrt{130}}{100} = 0.01 \times 11.4 = 0.114$
Now,add these values together:
$1.14 + 36.05 + 0.114 = 37.304$
Thus,the value is $37.304$.

(C) To find the square root of $4356$,we perform prime factorization:
$4356 = 2 \times 2178$
$2178 = 2 \times 1089$
$1089 = 3 \times 363$
$363 = 3 \times 121$
$121 = 11 \times 11$
Thus,the prime factorization is $4356 = 2^2 \times 3^2 \times 11^2$.
Taking the square root of both sides:
$\sqrt{4356} = \sqrt{2^2 \times 3^2 \times 11^2}$
$\sqrt{4356} = 2 \times 3 \times 11$
$\sqrt{4356} = 66$
Therefore,the correct option is $C$.

(A) To find the square root of $104976$,we can use the prime factorization method or the long division method.
Step $1$: Pair the digits from right to left: $10, 49, 76$.
Step $2$: Find the largest number whose square is less than or equal to $10$. That is $3^2 = 9$.
Step $3$: Subtract $9$ from $10$ to get $1$,then bring down $49$ to get $149$.
Step $4$: Double the quotient $(3 \times 2 = 6)$ and find a digit $x$ such that $6x \times x \leq 149$. For $x = 2$,$62 \times 2 = 124$.
Step $5$: Subtract $124$ from $149$ to get $25$,then bring down $76$ to get $2576$.
Step $6$: Double the current quotient $(32 \times 2 = 64)$ and find a digit $y$ such that $64y \times y = 2576$. For $y = 4$,$644 \times 4 = 2576$.
Therefore,$\sqrt{104976} = 324$.

(A) To find the square root of $211600$,we can write it as: $\sqrt{211600} = \sqrt{2116 \times 100}$.
Since $\sqrt{100} = 10$,we need to find the square root of $2116$.
By prime factorization of $2116$: $2116 = 2^2 \times 23^2$.
Therefore,$\sqrt{2116} = 2 \times 23 = 46$.
Thus,$\sqrt{211600} = 46 \times 10 = 460$.

(C) To find the square root of $6492304$,we can use the long division method:
$1$. Group the digits in pairs from right to left: $6, 49, 23, 04$.
$2$. Find the largest square less than or equal to $6$,which is $2^2 = 4$. Subtract $4$ from $6$ to get $2$. Bring down $49$,making it $249$.
$3$. Double the divisor $(2 \times 2 = 4)$. Find a digit $x$ such that $4x \times x \leq 249$. For $x = 5$,$45 \times 5 = 225$. Subtract $225$ from $249$ to get $24$. Bring down $23$,making it $2423$.
$4$. Double the current divisor $(25 \times 2 = 50)$. Find a digit $y$ such that $50y \times y \leq 2423$. For $y = 4$,$504 \times 4 = 2016$. Subtract $2016$ from $2423$ to get $407$. Bring down $04$,making it $40704$.
$5$. Double the current divisor $(254 \times 2 = 508)$. Find a digit $z$ such that $508z \times z = 40704$. For $z = 8$,$5088 \times 8 = 40704$.
Therefore,$\sqrt{6492304} = 2548$.

(A) First,find the prime factorization of $74088$:
$74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7$
Grouping the prime factors into pairs:
$74088 = (2 \times 2) \times (3 \times 3) \times (7 \times 7) \times (2 \times 3 \times 7)$
To make the number a perfect square,every prime factor must be in a pair. The factors $2, 3,$ and $7$ are left unpaired.
Therefore,the required number to multiply is $2 \times 3 \times 7 = 42$.

$2$	$74088$
$2$	$37044$
$2$	$18522$
$3$	$9261$
$3$	$3087$
$3$	$1029$
$7$	$343$
$7$	$49$
$7$	$7$
	$1$

(D) Using the property of square roots,$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.
Therefore,$\sqrt{10} \times \sqrt{250} = \sqrt{10 \times 250}$.
$\sqrt{10 \times 250} = \sqrt{2500}$.
Since $50 \times 50 = 2500$,the square root of $2500$ is $50$.

(A) First,simplify each square root term by finding the prime factors:
$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$
$3\sqrt{245} = 3\sqrt{49 \times 5} = 3 \times 7\sqrt{5} = 21\sqrt{5}$
$\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$
Now,substitute these values back into the expression:
$4\sqrt{5} + 21\sqrt{5} - 5\sqrt{5}$
$= (4 + 21 - 5)\sqrt{5}$
$= 20\sqrt{5}$

(C) Let the missing value be $x$.
Then,the equation is $\frac{250}{\sqrt{x}} = 10$.
By cross-multiplying,we get $\sqrt{x} = \frac{250}{10}$.
$\sqrt{x} = 25$.
Squaring both sides,we get $x = (25)^2$.
Therefore,$x = 625$.

(A) Given the equation: $\frac{\sqrt{256}}{\sqrt{x}}=2$
Since $\sqrt{256} = 16$,we substitute this into the equation:
$\frac{16}{\sqrt{x}} = 2$
Multiply both sides by $\sqrt{x}$:
$16 = 2\sqrt{x}$
Divide both sides by $2$:
$8 = \sqrt{x}$
Square both sides to solve for $x$:
$x = 8^2 = 64$

(C) To find the smallest number by which $216$ should be divided to make it a perfect square,we first find the prime factorization of $216$.
$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2 \times 2) \times (3 \times 3) \times 2 \times 3 = 2^2 \times 3^2 \times 6$.
In the prime factorization,the factors $2$ and $3$ are in pairs,but the factor $6$ (which is $2 \times 3$) is left unpaired.
To make the number a perfect square,we must divide $216$ by this remaining factor,which is $6$.
Thus,$\frac{216}{6} = 2^2 \times 3^2 = 4 \times 9 = 36$,which is a perfect square $(6^2 = 36)$.
Therefore,the smallest number is $6$.

(C) Let the missing value be $x$.
Given the equation: $\frac{\sqrt{x}}{200} = 0.02$.
Multiply both sides by $200$:
$\sqrt{x} = 200 \times 0.02$
$\sqrt{x} = 4$.
Squaring both sides:
$x = 4^2 = 16$.

(C) To solve the expression $\frac{\sqrt{6727}}{\sqrt{7}}$,we can use the property of square roots: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$.
Applying this property,we get $\sqrt{\frac{6727}{7}}$.
Dividing $6727$ by $7$,we get $6727 \div 7 = 961$.
Therefore,the expression becomes $\sqrt{961}$.
Since $31 \times 31 = 961$,the square root of $961$ is $31$.

(A) To find the square root of $0.09$,we can express the decimal as a fraction:
$\sqrt{0.09} = \sqrt{\frac{9}{100}}$
Since $\sqrt{9} = 3$ and $\sqrt{100} = 10$,we have:
$\frac{3}{10} = 0.3$
Therefore,the correct option is $A$.

(A) To solve $\frac{14}{3+\sqrt{2}}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(3-\sqrt{2})$.
$\frac{14}{3+\sqrt{2}} = \frac{14(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$
Using the identity $(a+b)(a-b) = a^2 - b^2$,we get:
$= \frac{14(3-\sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3-\sqrt{2})}{9-2} = \frac{14(3-\sqrt{2})}{7}$
$= 2(3-\sqrt{2})$
Given $\sqrt{2} \approx 1.414$,we substitute the value:
$= 2(3 - 1.414) = 2(1.586) = 3.172$

(C) To find the smallest number that must be added to $3579$ to make it a perfect square,we first find the square root of $3579$ by long division method.
$\sqrt{3579} \approx 59.82$.
The next perfect square after $3579$ is the square of the next integer,which is $60^2$.
$60^2 = 3600$.
To find the required number,we subtract $3579$ from $3600$.
Required number $= 3600 - 3579 = 21$.

(A) Given the equation: $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13}$
First,simplify the expression inside the square root:
$1+\frac{27}{169} = \frac{169+27}{169} = \frac{196}{169}$
Now,take the square root of the fraction:
$\sqrt{\frac{196}{169}} = \frac{14}{13}$
Substitute this back into the original equation:
$\frac{14}{13} = 1+\frac{x}{13}$
Subtract $1$ from both sides:
$\frac{14}{13} - 1 = \frac{x}{13}$
$\frac{14-13}{13} = \frac{x}{13}$
$\frac{1}{13} = \frac{x}{13}$
Therefore,$x = 1$.

(C) To solve the expression $\frac{\sqrt{4375}}{\sqrt{7}}$,we use the property of radicals: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$.
Applying this property,we get $\sqrt{\frac{4375}{7}}$.
Dividing $4375$ by $7$,we get $4375 \div 7 = 625$.
Therefore,the expression becomes $\sqrt{625}$.
Since $25 \times 25 = 625$,the square root of $625$ is $25$.

(A) Given the equation: $\sqrt{0.04 \times 0.4 \times a} = 0.4 \times 0.04 \times \sqrt{b}$
First,calculate the product inside the square root on the left side: $0.04 \times 0.4 = 0.016$.
So,the equation becomes: $\sqrt{0.016 \times a} = 0.016 \times \sqrt{b}$.
Divide both sides by $\sqrt{b}$ and $\sqrt{0.016}$ to isolate the ratio $\frac{a}{b}$:
$\frac{\sqrt{a}}{\sqrt{b}} = \frac{0.016}{\sqrt{0.016}}$
$\sqrt{\frac{a}{b}} = \sqrt{0.016}$
Squaring both sides,we get: $\frac{a}{b} = 0.016$.

(D) Let the missing number be $x$.
Then,the equation is $\frac{\sqrt{1296}}{x} = \frac{x}{2.25}$.
We know that $\sqrt{1296} = 36$.
Substituting this value,we get $\frac{36}{x} = \frac{x}{2.25}$.
Cross-multiplying,we get $x^2 = 36 \times 2.25$.
$x^2 = 36 \times \frac{225}{100}$.
Taking the square root on both sides,$x = \sqrt{36 \times \frac{225}{100}}$.
$x = 6 \times \frac{15}{10} = 6 \times 1.5 = 9$.
Therefore,the missing number is $9$.

(B) First,find the square root of $2401$. Since $40^2 = 1600$ and $50^2 = 2500$,the number ends in $1$,so it must be $41$ or $49$. Checking $49^2$: $49 \times 49 = 2401$.
Therefore,$\sqrt{2401} = 49$.
Now,substitute this value into the expression: $\sqrt{176 + 49}$.
Calculate the sum: $176 + 49 = 225$.
Finally,find the square root of $225$: $\sqrt{225} = 15$.

(A) To solve $\sqrt{10} \times \sqrt{15}$,we use the property $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.
$\sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15} = \sqrt{150}$.
Now,factorize $150$ to simplify the square root:
$150 = 25 \times 6$.
Therefore,$\sqrt{150} = \sqrt{25 \times 6} = \sqrt{25} \times \sqrt{6}$.
Since $\sqrt{25} = 5$,the expression simplifies to $5 \sqrt{6}$.

(A) To solve the expression $\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}$,we first simplify the square roots of the individual terms.
$\sqrt{\frac{4}{3}} = \frac{\sqrt{4}}{\sqrt{3}} = \frac{2}{\sqrt{3}}$
$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$
Now,subtract the two fractions:
$\frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{2 \times 2 - \sqrt{3} \times \sqrt{3}}{2 \sqrt{3}}$
$= \frac{4 - 3}{2 \sqrt{3}}$
$= \frac{1}{2 \sqrt{3}}$

(B) To solve the expression $\sqrt{248+\sqrt{52+\sqrt{144}}}$,we start from the innermost square root.
First,calculate $\sqrt{144} = 12$.
Substitute this value into the expression: $\sqrt{248+\sqrt{52+12}}$.
Next,simplify the term inside the square root: $52 + 12 = 64$.
Now,calculate $\sqrt{64} = 8$.
Substitute this back: $\sqrt{248+8}$.
Finally,calculate $\sqrt{256} = 16$.

(B) Given expression is $\frac{\sqrt{0.0009}}{\sqrt{0.01}}$.
Using the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$,we get $\sqrt{\frac{0.0009}{0.01}}$.
Multiplying numerator and denominator by $10000$ to remove decimals,we get $\sqrt{\frac{9}{100}}$.
Calculating the square root,$\sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3$.

(D) To solve the expression $\frac{1}{\sqrt{9}-\sqrt{8}}$,we rationalize the denominator.
First,simplify the square roots: $\sqrt{9} = 3$ and $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.
So,the expression becomes $\frac{1}{3-2\sqrt{2}}$.
Now,multiply the numerator and the denominator by the conjugate $(3+2\sqrt{2})$:
$\frac{1}{3-2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}$.
Calculate the denominator: $(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1$.
Therefore,the final result is $\frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}$.

(B) Given the equation: $\sqrt{\frac{x}{169}} = \frac{54}{39}$
Square both sides of the equation:
$\frac{x}{169} = \left(\frac{54}{39}\right)^2$
$\frac{x}{169} = \frac{54 \times 54}{39 \times 39}$
Since $39^2 = 1521$ and $169 = 13^2$,we can simplify the expression:
$x = \frac{54 \times 54}{39 \times 39} \times 169$
Note that $169 = 13 \times 13$ and $39 = 3 \times 13$,so $\frac{169}{39 \times 39} = \frac{169}{1521} = \frac{1}{9}$.
Therefore,$x = \frac{54 \times 54}{9} = 6 \times 54 = 324$.

(B) Let the given expression be $x = \sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}$.
Squaring both sides,we get $x^2 = 12 + \sqrt{12+\sqrt{12+\cdots}}$.
Since the expression is infinite,the term under the square root is again $x$,so $x^2 = 12 + x$.
Rearranging the terms,we get the quadratic equation $x^2 - x - 12 = 0$.
Factoring the quadratic equation,we get $(x - 4)(x + 3) = 0$.
This gives two possible values for $x$: $x = 4$ or $x = -3$.
Since the square root of a positive number must be positive,we neglect $x = -3$.
Therefore,$x = 4$.

(D) First,evaluate the square roots in the expression:
$\sqrt{196} = 14$
$\sqrt{576} = 24$
$\sqrt{256} = 16$
Now,substitute these values into the expression:
$= \frac{112}{14} \times \frac{24}{12} \times \frac{16}{8}$
$= 8 \times 2 \times 2$
$= 32$

(C) Given: $\sqrt{12} = 3.464$.
We need to evaluate $\sqrt{\frac{3}{4}} + 2 \sqrt{\frac{4}{3}}$.
Simplify the terms:
$\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} = \frac{\sqrt{3} \times 2}{2 \times 2} = \frac{\sqrt{12}}{4}$.
$2 \sqrt{\frac{4}{3}} = 2 \times \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4 \times \sqrt{3}}{3} = \frac{4 \times \sqrt{3} \times 4}{3 \times 4} = \frac{16 \sqrt{3}}{12} = \frac{8 \sqrt{12}}{12} = \frac{2 \sqrt{12}}{3}$.
Adding them together:
$\frac{\sqrt{12}}{4} + \frac{2 \sqrt{12}}{3} = \sqrt{12} \left( \frac{1}{4} + \frac{2}{3} \right) = \sqrt{12} \left( \frac{3 + 8}{12} \right) = \sqrt{12} \left( \frac{11}{12} \right)$.
Substitute $\sqrt{12} = 3.464$:
$= 3.464 \times \frac{11}{12} = \frac{38.104}{12} = 3.175333... \approx 3.1753$.

(C) Given that $\sqrt{15625} = 125$.
We need to evaluate the expression: $\sqrt{15625} + \sqrt{156.25} + \sqrt{1.5625}$.
First,express the decimals as fractions:
$\sqrt{156.25} = \sqrt{\frac{15625}{100}} = \frac{\sqrt{15625}}{\sqrt{100}} = \frac{125}{10} = 12.5$.
$\sqrt{1.5625} = \sqrt{\frac{15625}{10000}} = \frac{\sqrt{15625}}{\sqrt{10000}} = \frac{125}{100} = 1.25$.
Now,substitute these values back into the original expression:
$125 + 12.5 + 1.25 = 138.75$.

(A) Given equation: $\sqrt{0.03 \times 0.3 \times a} = 0.03 \times 0.3 \times \sqrt{b}$
Divide both sides by $\sqrt{b}$:
$\frac{\sqrt{0.03 \times 0.3 \times a}}{\sqrt{b}} = 0.03 \times 0.3$
Combine the square roots on the left side:
$\sqrt{\frac{0.03 \times 0.3 \times a}{b}} = 0.03 \times 0.3$
Square both sides to remove the square root:
$\frac{0.03 \times 0.3 \times a}{b} = (0.03 \times 0.3)^2$
Divide both sides by $(0.03 \times 0.3)$:
$\frac{a}{b} = 0.03 \times 0.3$
Calculate the product:
$\frac{a}{b} = 0.009$

(B) We are given $\sqrt{4096} = 64$.
We need to evaluate the expression: $\sqrt{4096} + \sqrt{40.96} + \sqrt{0.004096}$.
Step $1$: Simplify each term.
$\sqrt{4096} = 64$
$\sqrt{40.96} = \sqrt{\frac{4096}{100}} = \frac{64}{10} = 6.4$
$\sqrt{0.004096} = \sqrt{\frac{4096}{1000000}} = \frac{64}{1000} = 0.064$
Step $2$: Add the values together.
$64 + 6.4 + 0.064 = 70.464$.

(B) Given the equation: $\sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7}$
First,simplify the term inside the inner square root: $1-\frac{2176}{2401} = \frac{2401-2176}{2401} = \frac{225}{2401}$
Now,take the square root of this fraction: $\sqrt{\frac{225}{2401}} = \frac{15}{49}$
Substitute this back into the original equation: $\sqrt{1+\frac{15}{49}} = 1+\frac{x}{7}$
Simplify the expression inside the outer square root: $\sqrt{\frac{49+15}{49}} = \sqrt{\frac{64}{49}}$
Calculate the square root: $\frac{8}{7} = 1+\frac{x}{7}$
Rewrite $\frac{8}{7}$ as $1+\frac{1}{7}$: $1+\frac{1}{7} = 1+\frac{x}{7}$
Comparing both sides,we get $x = 1$.

(A) $1$. The square of an odd number must end in an odd digit $(1, 5, 9)$. Option $C$ ends in $4$,which is even,so it is eliminated.
$2$. $A$ $3$-digit number $n$ ranges from $100$ to $999$. The square $n^2$ ranges from $100^2 = 10,000$ to $999^2 = 998,001$.
$3$. Therefore,the square of a $3$-digit number must have either $5$ or $6$ digits.
$4$. Option $A$ $(65xxx1)$ has $6$ digits,which is possible.
$5$. Option $B$ $(9xx1)$ has only $4$ digits,which is impossible for a $3$-digit number.
$6$. Option $D$ $(9xxxxxx5)$ has $8$ digits,which is impossible for a $3$-digit number.
$7$. Thus,$65xxx1$ is the only possible candidate.

(C) Given expression $= \sqrt{\frac{0.324 \times 0.081 \times 4.624}{1.5625 \times 0.0289 \times 72.9 \times 64}}$
First,count the decimal places in the numerator and denominator.
Numerator: $3 + 3 + 3 = 9$ decimal places.
Denominator: $4 + 4 + 1 + 0 = 9$ decimal places.
Since the number of decimal places is equal,we can remove the decimals:
$= \sqrt{\frac{324 \times 81 \times 4624}{15625 \times 289 \times 729 \times 64}}$
Now,calculate the square roots of the individual numbers:
$= \frac{\sqrt{324} \times \sqrt{81} \times \sqrt{4624}}{\sqrt{15625} \times \sqrt{289} \times \sqrt{729} \times \sqrt{64}}$
$= \frac{18 \times 9 \times 68}{125 \times 17 \times 27 \times 8}$
Simplify the fraction:
$= \frac{18 \times 9 \times 68}{125 \times 17 \times 27 \times 8} = \frac{11016}{459000} = \frac{3}{125} = 0.024$

(A) Given expression is $\sqrt{0.01+\sqrt{0.0064}}$.
First,evaluate the inner square root: $\sqrt{0.0064} = 0.08$.
Now,substitute this value back into the expression: $\sqrt{0.01 + 0.08}$.
Add the numbers inside the square root: $\sqrt{0.09}$.
Finally,calculate the square root: $\sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3$.

(C) To find the cube root of $0.000064$,we first express the decimal as a fraction:
$\sqrt[3]{0.000064} = \sqrt[3]{\frac{64}{1000000}}$
Next,we express $64$ as $4^3$ and $1000000$ as $100^3$:
$= \sqrt[3]{\frac{4^3}{100^3}}$
Taking the cube root:
$= \frac{4}{100} = 0.04$

(C) To find the least number by which $14175$ should be divided to make it a perfect square,we first find its prime factorization.
$14175 = 5 \times 2835 = 5 \times 5 \times 567 = 5 \times 5 \times 3 \times 189 = 5 \times 5 \times 3 \times 3 \times 63 = 5 \times 5 \times 3 \times 3 \times 3 \times 21 = 5 \times 5 \times 3 \times 3 \times 3 \times 3 \times 7$.
Thus,$14175 = 5^2 \times 3^4 \times 7$.
For a number to be a perfect square,the exponent of each prime factor must be even.
In the prime factorization of $14175$,the prime factor $7$ has an exponent of $1$,which is odd.
Therefore,to make the number a perfect square,we must divide $14175$ by $7$ to eliminate the factor $7^1$.

(B) To simplify the expression,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{5}-\sqrt{3})$.
$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
$= \frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5})^2-(\sqrt{3})^2}$
$= \frac{5 + 3 - 2\sqrt{15}}{5-3}$
$= \frac{8 - 2\sqrt{15}}{2}$
$= \frac{2(4 - \sqrt{15})}{2} = 4 - \sqrt{15}$

(C) To solve the expression $\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}$,we simplify each radical term by factoring out the perfect squares:
$\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$
$\sqrt{216} = \sqrt{36 \times 6} = 6\sqrt{6}$
$\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$
Now,substitute these values back into the expression:
$\frac{2\sqrt{6} + 6\sqrt{6}}{4\sqrt{6}} = \frac{(2+6)\sqrt{6}}{4\sqrt{6}} = \frac{8\sqrt{6}}{4\sqrt{6}}$
Canceling $\sqrt{6}$ from the numerator and denominator:
$\frac{8}{4} = 2$

(C) First,convert the mixed fraction $2 \frac{2}{9}$ into an improper fraction:
$2 \frac{2}{9} = \frac{2 \times 9 + 2}{9} = \frac{18 + 2}{9} = \frac{20}{9}$.
Now,find the square root:
$\sqrt{\frac{20}{9}} = \frac{\sqrt{20}}{\sqrt{9}}$.
Given that $\sqrt{20} = 4.472$ and $\sqrt{9} = 3$,substitute these values:
$\frac{4.472}{3} = 1.49066...$.
Rounding to two decimal places,we get $1.49$.

(B) First,rationalize $a$ and $b$:
$a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^{2}}{5 - 1} = \frac{5 + 1 + 2\sqrt{5}}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$
$b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{(\sqrt{5} - 1)^{2}}{5 - 1} = \frac{5 + 1 - 2\sqrt{5}}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$
Now,calculate $a + b$ and $ab$:
$a + b = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{6}{2} = 3$
$ab = \left(\frac{3 + \sqrt{5}}{2}\right) \left(\frac{3 - \sqrt{5}}{2}\right) = \frac{9 - 5}{4} = \frac{4}{4} = 1$
We know that $a^{2} + b^{2} = (a + b)^{2} - 2ab = (3)^{2} - 2(1) = 9 - 2 = 7$
Substitute these into the expression:
$\frac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \frac{(a^{2} + b^{2}) + ab}{(a^{2} + b^{2}) - ab} = \frac{7 + 1}{7 - 1} = \frac{8}{6} = \frac{4}{3}$

(C) To find the least number by which $10584$ must be multiplied to make it a perfect square,we first find its prime factorization.
$10584 = 2 \times 5292 = 2^2 \times 2646 = 2^3 \times 1323 = 2^3 \times 3 \times 441 = 2^3 \times 3 \times 3 \times 147 = 2^3 \times 3^2 \times 3 \times 49 = 2^3 \times 3^3 \times 7^2$.
Grouping the prime factors into pairs: $10584 = (2^2 \times 3^2 \times 7^2) \times (2 \times 3)$.
For a number to be a perfect square,all prime factors must have even exponents.
Here,the factors $2$ and $3$ have odd exponents ($1$ each).
Therefore,we must multiply by $2 \times 3 = 6$ to make all exponents even.

(B) First,find the prime factorization of $7936$.
$7936 = 2 \times 3968 = 2^2 \times 1984 = 2^3 \times 992 = 2^4 \times 496 = 2^5 \times 248 = 2^6 \times 124 = 2^7 \times 62 = 2^8 \times 31$.
For a number to be a perfect square,the exponent of each prime factor must be an even number.
In $7936 = 2^8 \times 31^1$,the exponent of $2$ is $8$ (which is even),but the exponent of $31$ is $1$ (which is odd).
To make it a perfect square,we must multiply $7936$ by $31$ so that the exponent of $31$ becomes $2$.
Therefore,the smallest perfect square is $7936 \times 31 = 246016$.

(B) To find the square root of $0.00059049$,we first express the decimal as a fraction:
$\sqrt{0.00059049} = \sqrt{\frac{59049}{100000000}}$
Next,we take the square root of the numerator and the denominator separately:
$= \frac{\sqrt{59049}}{\sqrt{100000000}}$
Since $243^2 = 59049$ and $\sqrt{100000000} = 10000$,we get:
$= \frac{243}{10000}$
$= 0.0243$

(C) To find the value of $\sqrt{\frac{4}{12.1}}$,we first simplify the expression inside the square root by multiplying the numerator and denominator by $10$ to remove the decimal:
$\sqrt{\frac{4}{12.1}} = \sqrt{\frac{4 \times 10}{12.1 \times 10}} = \sqrt{\frac{40}{121}}$
$= \frac{\sqrt{4} \times \sqrt{10}}{\sqrt{121}}$
$= \frac{2 \times 3.16}{11}$
$= \frac{6.32}{11}$
$= 0.5745...$
Rounding to one decimal place,we get $0.6$.

(D) To solve the expression $\sqrt{\frac{0.256 \times 0.081 \times 4.356}{1.5625 \times 0.0121 \times 129.6 \times 64}}$,first adjust the decimal points.
The numerator has $3 + 3 + 3 = 9$ decimal places,and the denominator has $4 + 4 + 1 + 0 = 9$ decimal places.
Since the number of decimal places is equal,we can remove the decimals:
$= \sqrt{\frac{256 \times 81 \times 4356}{15625 \times 121 \times 1296 \times 64}}$
Now,calculate the square roots of the individual numbers:
$= \frac{\sqrt{256} \times \sqrt{81} \times \sqrt{4356}}{\sqrt{15625} \times \sqrt{121} \times \sqrt{1296} \times \sqrt{64}}$
$= \frac{16 \times 9 \times 66}{125 \times 11 \times 36 \times 8}$
Simplify the fraction:
$= \frac{16 \times 9 \times 66}{125 \times 11 \times 36 \times 8} = \frac{144 \times 66}{1375 \times 288} = \frac{9504}{396000} = 0.024$

(A) Let the number of men in the front row be $x$. Since the men are arranged in a solid square, the total number of men used in the square is $x^2$.
Given that there are $16160$ men in total and $31$ men are left over, the number of men used in the square is $16160 - 31 = 16129$.
Therefore, $x^2 = 16129$.
To find $x$, we calculate the square root of $16129$: $x = \sqrt{16129} = 127$.
Thus, the number of men in the front row is $127$.