Simplification Questions in English

(C) To find the approximate value,we round the given numbers to the nearest whole numbers:
$25.05 \approx 25$
$123.95 \approx 124$
$388.999 \approx 389$
$15.001 \approx 15$
Now,substitute these values into the expression:
$25 \times 124 + 389 \times 15$
$= 3100 + 5835$
$= 8935$
Thus,the value in place of the question mark is $8935$.

(A) To solve the expression $561 \div 35.05 \times 19.99$,we use approximation.
$1$. Approximate $35.05$ to $35$.
$2$. Approximate $19.99$ to $20$.
$3$. The expression becomes: $561 \div 35 \times 20$.
$4$. Calculate the division: $561 \div 35 \approx 16.028$.
$5$. Multiply by $20$: $16.028 \times 20 = 320.56$.
$6$. Rounding to the nearest integer,we get $320$.

(D) Given equation: $(21)^{2} - 3717 \div 59 = ? \times 8$
First,calculate the square of $21$: $(21)^{2} = 441$.
Next,perform the division: $3717 \div 59 = 63$.
Substitute these values back into the equation: $441 - 63 = ? \times 8$.
Subtract the values: $378 = ? \times 8$.
Solve for $?$: $? = \frac{378}{8} = 47.25$.

(D) To find the value of $?$,we rearrange the equation:
$? = 2 \frac{1}{8} - 1 \frac{1}{16} - 1 \frac{1}{32} + 1 \frac{9}{64}$
Separate the whole numbers and the fractions:
$? = (2 - 1 - 1 + 1) + (\frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{9}{64})$
Calculate the whole number part:
$2 - 1 - 1 + 1 = 1$
Calculate the fractional part by finding the common denominator,which is $64$:
$\frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{9}{64} = \frac{8}{64} - \frac{4}{64} - \frac{2}{64} + \frac{9}{64}$
$= \frac{8 - 4 - 2 + 9}{64} = \frac{11}{64}$
Combining both parts,we get:
$? = 1 + \frac{11}{64} = 1 \frac{11}{64}$

(C) Given equation: $(0.64)^{4} \div (0.512)^{3} \times (0.8)^{4} = (0.8)^{?+3}$
Express all terms with base $0.8$:
$0.64 = (0.8)^{2}$
$0.512 = (0.8)^{3}$
Substituting these values:
$[(0.8)^{2}]^{4} \div [(0.8)^{3}]^{3} \times (0.8)^{4} = (0.8)^{?+3}$
Using the power rule $(a^{m})^{n} = a^{m \times n}$:
$(0.8)^{8} \div (0.8)^{9} \times (0.8)^{4} = (0.8)^{?+3}$
Using the exponent rules $a^{m} \div a^{n} = a^{m-n}$ and $a^{m} \times a^{n} = a^{m+n}$:
$(0.8)^{8-9+4} = (0.8)^{?+3}$
$(0.8)^{3} = (0.8)^{?+3}$
Equating the exponents:
$3 = ? + 3$
$? = 3 - 3 = 0$

(D) To solve the expression $\sqrt{15^{2} \times 12 \div 9 - 125 + 21}$,follow the order of operations $(BODMAS)$:
$1$. Calculate the square: $15^{2} = 225$.
$2$. Perform division: $12 \div 9 = \frac{12}{9} = \frac{4}{3}$.
$3$. Perform multiplication: $225 \times \frac{4}{3} = 75 \times 4 = 300$.
$4$. Perform addition and subtraction: $300 - 125 + 21 = 175 + 21 = 196$.
$5$. Calculate the square root: $\sqrt{196} = 14$.
Therefore,the value is $14$.

(B) Given equation: $7441 \div 34 \times 12 = ? \times 9 + 110$
Step $1$: Perform the division and multiplication approximately.
$7441 \div 34 \approx 218.85$
$218.85 \times 12 \approx 2626.2$
Step $2$: Substitute the value into the equation.
$2626.2 = ? \times 9 + 110$
Step $3$: Subtract $110$ from both sides.
$2626.2 - 110 = ? \times 9$
$2516.2 = ? \times 9$
Step $4$: Divide by $9$ to find the value of $?$.
$? = 2516.2 / 9 \approx 279.57$
Rounding to the nearest whole number,we get $? \approx 280$.

(C) To find the approximate value,we simplify the expression:
$? = \frac{989}{34} \div \frac{65}{869} \times \frac{515}{207}$
$= \frac{989}{34} \times \frac{869}{65} \times \frac{515}{207}$
Approximating the values:
$\frac{989}{34} \approx \frac{990}{33} = 30$
$\frac{869}{65} \approx \frac{870}{65} \approx 13.38$
$\frac{515}{207} \approx \frac{517.5}{207} = 2.5$
Calculation: $30 \times 13.38 \times 2.5 \approx 1003.5$
Alternatively,calculating directly: $\frac{989}{34} \approx 29.08$,$\frac{869}{65} \approx 13.37$,$\frac{515}{207} \approx 2.48$
$29.08 \times 13.37 \times 2.48 \approx 964.3$
Rounding to the nearest given option,the value is approximately $970$.

(D) To find the approximate value,we round the numbers to the nearest integers:
$(32.13) \approx 32$
$(23.96) \approx 24$
$(17.11) \approx 17$
Now,substitute these values into the expression:
$? = (32)^{2} + (24)^{2} - (17)^{2}$
Calculate the squares:
$? = 1024 + 576 - 289$
Perform the addition and subtraction:
$? = 1600 - 289$
$? = 1311$
The closest approximate value among the given options is $1310$.

(D) To solve the expression $4-\frac{5}{1+\frac{1}{3+\frac{1}{2+\frac{1}{4}}}}$,we simplify from the bottom up.
First,simplify the lowest fraction: $2+\frac{1}{4} = \frac{8+1}{4} = \frac{9}{4}$.
Next,substitute this back: $3+\frac{1}{9/4} = 3+\frac{4}{9} = \frac{27+4}{9} = \frac{31}{9}$.
Then,substitute this back: $1+\frac{1}{31/9} = 1+\frac{9}{31} = \frac{31+9}{31} = \frac{40}{31}$.
Finally,substitute this into the main expression: $4-\frac{5}{40/31} = 4-\frac{5 \times 31}{40} = 4-\frac{155}{40} = 4-\frac{31}{8}$.
Calculating the final value: $\frac{32-31}{8} = \frac{1}{8}$.

(C) Let $x = 1 . \overline{27} = 1.272727...$ (Equation $1$)
Multiply both sides by $100$ since two digits are repeating:
$100x = 127.272727...$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$100x - x = 127.272727... - 1.272727...$
$99x = 126$
$x = \frac{126}{99}$
Divide both numerator and denominator by $9$:
$x = \frac{14}{11}$

(B) Given equation: $2p + \frac{1}{p} = 4$.
Divide the entire equation by $2$:
$p + \frac{1}{2p} = 2$.
Now,cube both sides of the equation:
$\left(p + \frac{1}{2p}\right)^{3} = 2^{3}$.
Using the identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:
$p^{3} + \left(\frac{1}{2p}\right)^{3} + 3(p)\left(\frac{1}{2p}\right)\left(p + \frac{1}{2p}\right) = 8$.
$p^{3} + \frac{1}{8p^{3}} + \frac{3}{2}\left(p + \frac{1}{2p}\right) = 8$.
Substitute the value $p + \frac{1}{2p} = 2$ into the equation:
$p^{3} + \frac{1}{8p^{3}} + \frac{3}{2}(2) = 8$.
$p^{3} + \frac{1}{8p^{3}} + 3 = 8$.
$p^{3} + \frac{1}{8p^{3}} = 8 - 3 = 5$.

(D) To solve the expression $(0.1 \times 0.01 \times 0.001 \times 10^{7})$,we first convert the decimals into powers of $10$.
$0.1 = 10^{-1}$
$0.01 = 10^{-2}$
$0.001 = 10^{-3}$
Now,substitute these values into the expression:
$10^{-1} \times 10^{-2} \times 10^{-3} \times 10^{7}$
Using the law of exponents $a^{m} \times a^{n} = a^{m+n}$,we add the exponents:
$10^{(-1 + -2 + -3 + 7)} = 10^{(-6 + 7)} = 10^{1} = 10$.

(C) Given expression: $\left[\left(\sqrt[5]{x^{-3 / 5}}\right)^{-5 / 3}\right]^{5}$
Using the property $\sqrt[n]{a} = a^{1/n}$,we can write $\sqrt[5]{x^{-3/5}} = (x^{-3/5})^{1/5} = x^{-3/25}$.
Now,applying the power rule $(a^m)^n = a^{m \times n}$ repeatedly:
$= \left( (x^{-3/5})^{1/5} \right)^{-5/3 \times 5}$
$= (x^{-3/5})^{1/5 \times -5/3 \times 5}$
$= (x^{-3/5})^{-25/15}$
$= (x^{-3/5})^{-5/3}$
$= x^{(-3/5) \times (-5/3)}$
$= x^{1} = x$.

(A) The given expression is $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots \left(1-\frac{1}{25}\right)$.
Simplifying each term:
$= \left(\frac{3-1}{3}\right) \times \left(\frac{4-1}{4}\right) \times \left(\frac{5-1}{5}\right) \times \ldots \times \left(\frac{25-1}{25}\right)$
$= \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \ldots \times \frac{23}{24} \times \frac{24}{25}$
Observing the pattern,the numerator of each fraction cancels out the denominator of the preceding fraction.
$= \frac{2}{{3}} \times \frac{{3}}{{4}} \times \frac{{4}}{{5}} \times \ldots \times \frac{{24}}{25}$
$= \frac{2}{25}$

(B) Given equation: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2$
Rationalizing the denominator:
$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})} = 2$
$\frac{(3+x) + (3-x) + 2\sqrt{(3+x)(3-x)}}{(3+x) - (3-x)} = 2$
$\frac{6 + 2\sqrt{9-x^2}}{2x} = 2$
$\frac{2(3 + \sqrt{9-x^2})}{2x} = 2$
$3 + \sqrt{9-x^2} = 2x$
$\sqrt{9-x^2} = 2x - 3$
Squaring both sides:
$9 - x^2 = (2x - 3)^2$
$9 - x^2 = 4x^2 - 12x + 9$
$5x^2 - 12x = 0$
$x(5x - 12) = 0$
Since $x$ must satisfy the original equation,$x = \frac{12}{5}$ (as $x=0$ leads to $1=2$ which is false).

(C) Let $x = 0.121212 \ldots$ (Equation $1$)
Since there are two repeating digits,multiply both sides by $100$:
$100x = 12.121212 \ldots$ (Equation $2$)
Subtract Equation $1$ from Equation $2$:
$100x - x = 12.121212 \ldots - 0.121212 \ldots$
$99x = 12$
$x = \frac{12}{99}$
Dividing both numerator and denominator by their greatest common divisor,which is $3$:
$x = \frac{12 \div 3}{99 \div 3} = \frac{4}{33}$

(B) First,convert the mixed fractions into improper fractions:
$3 \frac{3}{4} = \frac{15}{4}$,$4 \frac{2}{5} = \frac{22}{5}$,and $3 \frac{1}{8} = \frac{25}{8}$.
Now,the expression becomes: $\frac{15}{4} + \frac{22}{5} - \frac{25}{8}$.
The least common multiple $(LCM)$ of the denominators $4, 5, \text{ and } 8$ is $40$.
Convert each fraction to have a denominator of $40$:
$\frac{15 \times 10}{4 \times 10} = \frac{150}{40}$,$\frac{22 \times 8}{5 \times 8} = \frac{176}{40}$,and $\frac{25 \times 5}{8 \times 5} = \frac{125}{40}$.
Now,perform the addition and subtraction:
$\frac{150 + 176 - 125}{40} = \frac{326 - 125}{40} = \frac{201}{40}$.
Finally,convert $\frac{201}{40}$ back into a mixed fraction:
$201 \div 40 = 5$ with a remainder of $1$,so $\frac{201}{40} = 5 \frac{1}{40}$.

(D) Given equation: $\sqrt{5^{2} \times 14 - 6 \times 7 + (4)^{?}} = 18$
Squaring both sides:
$5^{2} \times 14 - 6 \times 7 + (4)^{?} = (18)^{2}$
Calculating the values:
$25 \times 14 - 42 + (4)^{?} = 324$
$350 - 42 + (4)^{?} = 324$
$308 + (4)^{?} = 324$
Isolating the term with the unknown:
$(4)^{?} = 324 - 308$
$(4)^{?} = 16$
Expressing $16$ as a power of $4$:
$(4)^{?} = 4^{2}$
Therefore,$? = 2$. Since $2$ is not among the options $A, B, C$,the correct answer is $D$.

(A) Given: $x^{1/3} + y^{1/3} = z^{1/3}$ ......$(1)$
Cube both sides of the equation:
$(x^{1/3} + y^{1/3})^3 = (z^{1/3})^3$
Using the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$:
$x + y + 3x^{1/3}y^{1/3}(x^{1/3} + y^{1/3}) = z$
Substitute $x^{1/3} + y^{1/3} = z^{1/3}$ from equation $(1)$:
$x + y + 3x^{1/3}y^{1/3}z^{1/3} = z$
Rearrange the terms:
$x + y - z = -3x^{1/3}y^{1/3}z^{1/3}$
Cube both sides again:
$(x + y - z)^3 = (-3x^{1/3}y^{1/3}z^{1/3})^3$
$(x + y - z)^3 = -27xyz$
Therefore:
$(x + y - z)^3 + 27xyz = 0$

(C) Let $x = \sqrt{7 \sqrt{7 \sqrt{7 \cdots}}}$.
Squaring both sides,we get $x^2 = 7 \sqrt{7 \sqrt{7 \cdots}} = 7x$.
Since $x \neq 0$,we have $x = 7$.
Given the equation $(343)^{y-1} = 7$.
We know that $343 = 7^3$,so $(7^3)^{y-1} = 7^1$.
This simplifies to $7^{3(y-1)} = 7^1$.
Equating the exponents,$3(y-1) = 1$.
$3y - 3 = 1 \Rightarrow 3y = 4$.
Therefore,$y = \frac{4}{3}$.

(C) Given equations are $a+b+c=1$ and $ab+bc+ca=\frac{1}{3}$.
We know the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.
Substituting the given values: $(1)^2 = a^2+b^2+c^2+2(\frac{1}{3})$.
$1 = a^2+b^2+c^2+\frac{2}{3} \Rightarrow a^2+b^2+c^2 = 1-\frac{2}{3} = \frac{1}{3}$.
Now,consider the expression $(a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2) - 2(ab+bc+ca)$.
Substituting the values: $2(\frac{1}{3}) - 2(\frac{1}{3}) = 0$.
Since the sum of squares is zero,each term must be zero: $a-b=0, b-c=0, c-a=0$.
Thus,$a=b=c$.
Since $a+b+c=1$,we have $3a=1$,so $a=b=c=\frac{1}{3}$.
Therefore,the ratio $a:b:c = \frac{1}{3}:\frac{1}{3}:\frac{1}{3} = 1:1:1$.

(C) Given the equation: $a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}=4$
We can rearrange the terms as follows:
$(a^{2}-2+\frac{1}{a^{2}})+(b^{2}-2+\frac{1}{b^{2}})=0$
This can be written as:
$(a-\frac{1}{a})^{2}+(b-\frac{1}{b})^{2}=0$
Since the sum of squares of real numbers is zero only if each individual term is zero:
$(a-\frac{1}{a})=0 \Rightarrow a^{2}=1$
$(b-\frac{1}{b})=0 \Rightarrow b^{2}=1$
Therefore,the value of $a^{2}+b^{2} = 1+1 = 2$.

(D) Given that $(x + \frac{1}{x})^2 = 3$.
Taking the square root on both sides,we get $(x + \frac{1}{x}) = \sqrt{3}$.
We know the algebraic identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$.
Substituting $a = x$ and $b = \frac{1}{x}$:
$(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x)(\frac{1}{x})(x + \frac{1}{x})$.
Substituting the value $(x + \frac{1}{x}) = \sqrt{3}$:
$(\sqrt{3})^3 = x^3 + \frac{1}{x^3} + 3(1)(\sqrt{3})$.
$3\sqrt{3} = x^3 + \frac{1}{x^3} + 3\sqrt{3}$.
Subtracting $3\sqrt{3}$ from both sides:
$x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} = 0$.

(A) Let $x = 0.1$ and $y = 0.02$.
The given expression is $\frac{x^3 + y^3}{(2x)^3 + (2y)^3}$.
We know that $(2x)^3 = 8x^3$ and $(2y)^3 = 8y^3$.
Substituting these into the expression,we get $\frac{x^3 + y^3}{8x^3 + 8y^3}$.
Factoring out $8$ from the denominator,we get $\frac{x^3 + y^3}{8(x^3 + y^3)}$.
Canceling the common term $(x^3 + y^3)$,we are left with $\frac{1}{8}$.
Calculating the decimal value,$\frac{1}{8} = 0.125$.

(A) Given the equation $x+\frac{1}{x}=2$.
Multiplying both sides by $x$,we get $x^2+1=2x$,which simplifies to $x^2-2x+1=0$.
This is a perfect square: $(x-1)^2=0$,which implies $x=1$.
Now,substitute $x=1$ into the expression $x^{100}+\frac{1}{x^{100}}$:
$1^{100}+\frac{1}{1^{100}} = 1+1 = 2$.
Therefore,the value is $2$.

(C) Given equation: $x^{3}+3x^{2}+3x=7$
To solve for $x$,add $1$ to both sides of the equation to complete the cube:
$x^{3}+3x^{2}+3x+1 = 7+1$
Recognizing the algebraic identity $(a+b)^{3} = a^{3}+3a^{2}b+3ab^{2}+b^{3}$,we can rewrite the left side:
$(x+1)^{3} = 8$
Since $8 = 2^{3}$,we have:
$(x+1)^{3} = 2^{3}$
Taking the cube root on both sides:
$x+1 = 2$
Therefore:
$x = 2-1 = 1$

(B) Given that $2x + \frac{2}{x} = 1$.
Dividing the entire equation by $2$,we get $x + \frac{1}{x} = \frac{1}{2}$.
We know the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$.
Applying this to $x + \frac{1}{x}$,we have $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x + \frac{1}{x})$.
Substituting the value $x + \frac{1}{x} = \frac{1}{2}$:
$(\frac{1}{2})^{3} = x^{3} + \frac{1}{x^{3}} + 3(1)(\frac{1}{2})$.
$\frac{1}{8} = x^{3} + \frac{1}{x^{3}} + \frac{3}{2}$.
$x^{3} + \frac{1}{x^{3}} = \frac{1}{8} - \frac{3}{2}$.
$x^{3} + \frac{1}{x^{3}} = \frac{1 - 12}{8} = -\frac{11}{8}$.

(A) First,simplify the terms inside the square roots in the denominator.
$\sqrt{16+6 \sqrt{7}} = \sqrt{9+7+2 \times 3 \times \sqrt{7}} = \sqrt{3^2 + (\sqrt{7})^2 + 2 \times 3 \times \sqrt{7}} = \sqrt{(3+\sqrt{7})^2} = 3+\sqrt{7}$.
Similarly,$\sqrt{16-6 \sqrt{7}} = \sqrt{9+7-2 \times 3 \times \sqrt{7}} = \sqrt{(3-\sqrt{7})^2} = 3-\sqrt{7}$.
Now,substitute these values into the denominator:
$\sqrt{16+6 \sqrt{7}} - \sqrt{16-6 \sqrt{7}} = (3+\sqrt{7}) - (3-\sqrt{7}) = 3 + \sqrt{7} - 3 + \sqrt{7} = 2\sqrt{7}$.
Finally,substitute this back into the original expression:
$\frac{\sqrt{7}}{2\sqrt{7}} = \frac{1}{2}$.

(C) Given equation: $2x + \frac{1}{3x} = 6$
To find the value of $3x + \frac{1}{2x}$,we multiply the given equation by $\frac{3}{2}$:
$\frac{3}{2} \times (2x + \frac{1}{3x}) = \frac{3}{2} \times 6$
Distributing $\frac{3}{2}$ on the left side:
$(\frac{3}{2} \times 2x) + (\frac{3}{2} \times \frac{1}{3x}) = 9$
$3x + \frac{1}{2x} = 9$
Thus,the value is $9$.

(A) Given that $x=(\sqrt{2}-1)^{-1/2}$.
Squaring both sides,we get $x^2 = ((\sqrt{2}-1)^{-1/2})^2 = (\sqrt{2}-1)^{-1} = \frac{1}{\sqrt{2}-1}$.
Rationalizing the denominator for $x^2$: $x^2 = \frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$.
Now,find $\frac{1}{x^2}$: $\frac{1}{x^2} = \frac{1}{\sqrt{2}+1} = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.
Finally,calculate $x^2 - \frac{1}{x^2} = (\sqrt{2}+1) - (\sqrt{2}-1) = \sqrt{2} + 1 - \sqrt{2} + 1 = 2$.

(B) The given expression is $\frac{3}{4} \times (1+\frac{1}{3}) \times (1+\frac{2}{3}) \times (1-\frac{2}{5}) \times (1+\frac{6}{7}) \times (1-\frac{12}{13})$.
First,simplify each term inside the parentheses:
$(1+\frac{1}{3}) = \frac{4}{3}$
$(1+\frac{2}{3}) = \frac{5}{3}$
$(1-\frac{2}{5}) = \frac{3}{5}$
$(1+\frac{6}{7}) = \frac{13}{7}$
$(1-\frac{12}{13}) = \frac{1}{13}$
Now,substitute these values back into the expression:
$= \frac{3}{4} \times \frac{4}{3} \times \frac{5}{3} \times \frac{3}{5} \times \frac{13}{7} \times \frac{1}{13}$
Cancel out the common terms in the numerator and denominator:
$= (\frac{3}{4} \times \frac{4}{3}) \times (\frac{5}{3} \times \frac{3}{5}) \times (\frac{13}{7} \times \frac{1}{13})$
$= 1 \times 1 \times \frac{1}{7} = \frac{1}{7}$.

(C) Let $a = 0.87$ and $b = 0.13$.
The given expression is of the form $\frac{a^{3} + b^{3}}{a^{2} + b^{2} - ab}$.
Using the algebraic identity $a^{3} + b^{3} = (a + b)(a^{2} + b^{2} - ab)$,we can simplify the expression:
$\frac{(a + b)(a^{2} + b^{2} - ab)}{a^{2} + b^{2} - ab} = a + b$.
Substituting the values of $a$ and $b$ back into the simplified expression:
$0.87 + 0.13 = 1$.
Therefore,the value of the expression is $1$.

(D) Given equation: $x^{2}+y^{2}-2x+6y+10=0$
Rearranging the terms to complete the squares:
$(x^{2}-2x+1) + (y^{2}+6y+9) = 0$
This can be written as:
$(x-1)^{2} + (y+3)^{2} = 0$
Since the square of any real number is non-negative,the sum of two non-negative numbers can be zero only if each individual term is zero.
Therefore:
$(x-1)^{2} = 0 \implies x = 1$
$(y+3)^{2} = 0 \implies y = -3$
Now,calculate the value of $(x^{2}+y^{2})$:
$x^{2}+y^{2} = (1)^{2} + (-3)^{2}$
$x^{2}+y^{2} = 1 + 9 = 10$

(C) Given equation: $23 \times 15 - 60 + \frac{?}{31} = 292$
Step $1$: Calculate the product $23 \times 15 = 345$.
Step $2$: Substitute the value into the equation: $345 - 60 + \frac{?}{31} = 292$.
Step $3$: Simplify the expression: $285 + \frac{?}{31} = 292$.
Step $4$: Isolate the term with the question mark: $\frac{?}{31} = 292 - 285$.
Step $5$: Calculate the difference: $\frac{?}{31} = 7$.
Step $6$: Solve for $?$: $? = 7 \times 31 = 217$.

(B) To solve the expression $3 \frac{3}{4} + 4 \frac{2}{5} - 3 \frac{1}{8}$,we separate the whole numbers and the fractions:
$? = (3 + 4 - 3) + \left( \frac{3}{4} + \frac{2}{5} - \frac{1}{8} \right)$
First,solve the whole numbers: $(3 + 4 - 3) = 4$.
Next,find the least common multiple $(LCM)$ of the denominators $4, 5,$ and $8$,which is $40$.
Convert the fractions:
$\frac{3}{4} = \frac{30}{40}$
$\frac{2}{5} = \frac{16}{40}$
$\frac{1}{8} = \frac{5}{40}$
Now,combine the fractions:
$\frac{30 + 16 - 5}{40} = \frac{41}{40} = 1 \frac{1}{40}$
Finally,add the whole number part and the fraction part:
$4 + 1 \frac{1}{40} = 5 \frac{1}{40}$.

(A) To solve the expression $\frac{343 \times 49}{216 \times 16 \times 81}$,we express each number as a power of its prime factors:
$343 = 7^{3}$
$49 = 7^{2}$
$216 = 6^{3}$
$16 = 2^{4}$
$81 = 3^{4}$
Substituting these into the expression:
$\frac{7^{3} \times 7^{2}}{6^{3} \times 2^{4} \times 3^{4}}$
Since $2^{4} \times 3^{4} = (2 \times 3)^{4} = 6^{4}$,the expression becomes:
$\frac{7^{3+2}}{6^{3} \times 6^{4}} = \frac{7^{5}}{6^{3+4}} = \frac{7^{5}}{6^{7}}$

(C) Let the missing value be $x$.
Given equation: $6 \frac{1}{7} + 15 \frac{2}{3} + 11 \frac{1}{6} = 33 \frac{1}{21} - x$.
Convert mixed fractions to improper fractions:
$6 \frac{1}{7} = \frac{43}{7}$,$15 \frac{2}{3} = \frac{47}{3}$,$11 \frac{1}{6} = \frac{67}{6}$,$33 \frac{1}{21} = \frac{694}{21}$.
Sum of the left side: $\frac{43}{7} + \frac{47}{3} + \frac{67}{6}$.
The least common multiple of $7, 3, 6$ is $42$.
$= \frac{43 \times 6 + 47 \times 14 + 67 \times 7}{42} = \frac{258 + 658 + 469}{42} = \frac{1385}{42}$.
Now,$x = \frac{694}{21} - \frac{1385}{42}$.
$= \frac{1388 - 1385}{42} = \frac{3}{42} = \frac{1}{14}$.
Thus,the missing value is $\frac{1}{14}$.

(D) Given the equation: $57 \frac{4}{7} + 29 \frac{1}{21} + ? = 90 \frac{3}{35}$
Isolate the unknown term '$?$':
$? = 90 \frac{3}{35} - 57 \frac{4}{7} - 29 \frac{1}{21}$
Separate the whole numbers and fractions:
$? = (90 - 57 - 29) + (\frac{3}{35} - \frac{4}{7} - \frac{1}{21})$
Calculate the whole number part:
$90 - 57 - 29 = 4$
Find a common denominator for the fractions $(35, 7, 21)$,which is $105$:
$\frac{3}{35} = \frac{3 \times 3}{105} = \frac{9}{105}$
$\frac{4}{7} = \frac{4 \times 15}{105} = \frac{60}{105}$
$\frac{1}{21} = \frac{1 \times 5}{105} = \frac{5}{105}$
Combine the fractions:
$\frac{9}{105} - \frac{60}{105} - \frac{5}{105} = \frac{9 - 65}{105} = -\frac{56}{105}$
Simplify the expression:
$? = 4 - \frac{56}{105} = 3 + (1 - \frac{56}{105}) = 3 + \frac{105 - 56}{105} = 3 + \frac{49}{105}$
Reduce the fraction $\frac{49}{105}$ by dividing by $7$:
$\frac{49 \div 7}{105 \div 7} = \frac{7}{15}$
Thus,$? = 3 \frac{7}{15}$.

(C) Let the missing value be $x$.
$12 \frac{5}{7} + 23 \frac{3}{5} - 14 \frac{1}{3} = 28 \frac{4}{7} - x$
$x = 28 \frac{4}{7} + 14 \frac{1}{3} - 23 \frac{3}{5} - 12 \frac{5}{7}$
Group the whole numbers and fractions:
$x = (28 + 14 - 23 - 12) + (\frac{4}{7} + \frac{1}{3} - \frac{3}{5} - \frac{5}{7})$
$x = 7 + (\frac{4}{7} - \frac{5}{7} + \frac{1}{3} - \frac{3}{5})$
$x = 7 + (-\frac{1}{7} + \frac{5 - 9}{15})$
$x = 7 - \frac{1}{7} - \frac{4}{15}$
Find a common denominator for the fractions $(105)$:
$x = 7 - (\frac{15 + 28}{105})$
$x = 7 - \frac{43}{105}$
$x = 6 + (1 - \frac{43}{105}) = 6 + \frac{62}{105} = 6 \frac{62}{105}$

(A) Given expression: $7^{5} \div 7^{3} \div 7^{2} \times 7^{4} \times \frac{1}{7^{3}} = 7^{?}$
Using the laws of exponents,$a^{m} \div a^{n} = a^{m-n}$ and $a^{m} \times a^{n} = a^{m+n}$.
Also,$\frac{1}{7^{3}} = 7^{-3}$.
Substituting these into the expression:
$7^{5-3-2+4-3} = 7^{?}$
Calculating the exponent: $5 - 3 - 2 + 4 - 3 = 1$.
Therefore,$7^{1} = 7^{?}$.
Comparing the exponents,the value of $?$ is $1$.

(C) Let the missing value be $x$.
Given equation: $13 \frac{4}{x} + (5 \frac{3}{4} \text{ of } 3 \frac{1}{5}) \div 4 \frac{3}{5} = 17 \frac{4}{5}$
First,solve the 'of' operation: $5 \frac{3}{4} \text{ of } 3 \frac{1}{5} = \frac{23}{4} \times \frac{16}{5} = \frac{23 \times 4}{5} = \frac{92}{5} = 18 \frac{2}{5}$.
Now,perform the division: $(18 \frac{2}{5}) \div 4 \frac{3}{5} = \frac{92}{5} \div \frac{23}{5} = \frac{92}{5} \times \frac{5}{23} = 4$.
Substitute this back into the equation: $13 \frac{4}{x} + 4 = 17 \frac{4}{5}$.
Subtract $4$ from both sides: $13 \frac{4}{x} = 17 \frac{4}{5} - 4 = 13 \frac{4}{5}$.
Comparing the two sides,we get $x = 5$.

(B) Given equation: $47 \frac{1}{17} \div 1 \frac{49}{51} + 23 \frac{5}{7} + ? = 67 \frac{4}{9}$
Convert mixed fractions to improper fractions:
$\frac{800}{17} \div \frac{100}{51} + \frac{166}{7} + ? = \frac{607}{9}$
Perform the division:
$\frac{800}{17} \times \frac{51}{100} + \frac{166}{7} + ? = \frac{607}{9}$
$8 \times 3 + \frac{166}{7} + ? = \frac{607}{9}$
$24 + \frac{166}{7} + ? = \frac{607}{9}$
Isolate the variable $?$:
$? = \frac{607}{9} - 24 - \frac{166}{7}$
Find a common denominator $(63)$:
$? = \frac{607 \times 7 - 24 \times 63 - 166 \times 9}{63}$
$? = \frac{4249 - 1512 - 1494}{63} = \frac{1243}{63}$
Convert to mixed fraction:
$? = 19 \frac{46}{63}$

(C) Given equation: $22 \frac{2}{9} + 33 \frac{4}{7} - ? = 28 \frac{4}{45}$
Rearranging for $?$: $? = 22 \frac{2}{9} + 33 \frac{4}{7} - 28 \frac{4}{45}$
Separate the whole numbers and fractions: $? = (22 + 33 - 28) + (\frac{2}{9} + \frac{4}{7} - \frac{4}{45})$
Calculate the whole number part: $22 + 33 - 28 = 27$
Find the common denominator for the fractions $(9, 7, 45)$: The Least Common Multiple $(LCM)$ of $9, 7, 45$ is $315$.
Convert fractions: $\frac{2}{9} = \frac{2 \times 35}{315} = \frac{70}{315}$,$\frac{4}{7} = \frac{4 \times 45}{315} = \frac{180}{315}$,$\frac{4}{45} = \frac{4 \times 7}{315} = \frac{28}{315}$
Combine fractions: $\frac{70 + 180 - 28}{315} = \frac{222}{315}$
Simplify the fraction: $\frac{222 \div 3}{315 \div 3} = \frac{74}{105}$
Result: $27 + \frac{74}{105} = 27 \frac{74}{105}$

(C) First,separate the whole numbers and the fractions:
$1 + 1 + 1 + \frac{4}{7} + \frac{3}{5} + \frac{1}{3}$
$= 3 + \left( \frac{4}{7} + \frac{3}{5} + \frac{1}{3} \right)$
To add the fractions,find the least common multiple $(LCM)$ of $7, 5,$ and $3$,which is $105$.
$= 3 + \left( \frac{4 \times 15 + 3 \times 21 + 1 \times 35}{105} \right)$
$= 3 + \left( \frac{60 + 63 + 35}{105} \right)$
$= 3 + \frac{158}{105}$
Since $\frac{158}{105} = 1 \frac{53}{105}$,we have:
$= 3 + 1 + \frac{53}{105} = 4 \frac{53}{105}$

(A) To solve the expression $12 \frac{1}{3} + 10 \frac{5}{6} - 7 \frac{2}{3} - 1 \frac{4}{7}$,we separate the whole numbers and the fractions.
Step $1$: Separate the whole numbers and fractions:
$= (12 + 10 - 7 - 1) + (\frac{1}{3} + \frac{5}{6} - \frac{2}{3} - \frac{4}{7})$
Step $2$: Calculate the sum of the whole numbers:
$= 14 + (\frac{1}{3} + \frac{5}{6} - \frac{2}{3} - \frac{4}{7})$
Step $3$: Find a common denominator for the fractions $(3, 6, 7)$. The least common multiple is $42$:
$= 14 + (\frac{14}{42} + \frac{35}{42} - \frac{28}{42} - \frac{24}{42})$
Step $4$: Simplify the fraction part:
$= 14 + (\frac{14 + 35 - 28 - 24}{42}) = 14 + (\frac{-3}{42})$
Step $5$: Final calculation:
$= 14 - \frac{1}{14} = 13 + (1 - \frac{1}{14}) = 13 \frac{13}{14}$

(C) To solve the expression $4 \frac{3}{4} + 2 \frac{1}{8} + 7 \frac{1}{4} + 3 \frac{7}{8} + 11 \frac{12}{13}$,we group the whole numbers and the fractions separately.
First,group the terms:
$= (4 + 7 + 2 + 3 + 11) + (\frac{3}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{7}{8}) + \frac{12}{13}$
Calculate the sum of the whole numbers:
$4 + 7 + 2 + 3 + 11 = 27$
Calculate the sum of the fractions:
$(\frac{3}{4} + \frac{1}{4}) = \frac{4}{4} = 1$
$(\frac{1}{8} + \frac{7}{8}) = \frac{8}{8} = 1$
Add all the results together:
$= 27 + 1 + 1 + \frac{12}{13} = 29 \frac{12}{13}$

(A) To solve the expression $7 \frac{11}{18} + 13 \frac{1}{9} + 5 \frac{7}{9} - 16 \frac{2}{3}$,we separate the whole numbers and the fractions:
$= (7 + 13 + 5 - 16) + \left( \frac{11}{18} + \frac{1}{9} + \frac{7}{9} - \frac{2}{3} \right)$
First,calculate the sum of the whole numbers:
$7 + 13 + 5 - 16 = 9$
Next,find a common denominator for the fractions,which is $18$:
$= \frac{11}{18} + \frac{2}{18} + \frac{14}{18} - \frac{12}{18}$
$= \frac{11 + 2 + 14 - 12}{18} = \frac{15}{18}$
Simplify the fraction $\frac{15}{18}$ by dividing both numerator and denominator by $3$:
$= \frac{5}{6}$
Combining the whole number and the fraction,we get:
$= 9 \frac{5}{6}$

(A) Let the missing value be $x$.
$1 \frac{7}{9} + 3 \frac{5}{8} - 2 \frac{1}{18} + 4 \frac{7}{16} - 9 \frac{5}{18} + x = 0$
Separate the whole numbers and fractions:
$(1 + 3 - 2 + 4 - 9) + (\frac{7}{9} + \frac{5}{8} - \frac{1}{18} + \frac{7}{16} - \frac{5}{18}) + x = 0$
$-3 + (\frac{7}{9} - \frac{6}{18}) + \frac{5}{8} + \frac{7}{16} + x = 0$
$-3 + (\frac{14 - 6}{18}) + \frac{10 + 7}{16} + x = 0$
$-3 + \frac{8}{18} + \frac{17}{16} + x = 0$
$-3 + \frac{4}{9} + \frac{17}{16} + x = 0$
$-3 + \frac{64 + 153}{144} + x = 0$
$-3 + \frac{217}{144} + x = 0$
$-3 + 1 \frac{73}{144} + x = 0$
$-1 \frac{71}{144} + x = 0$
$x = 1 \frac{71}{144}$

(C) Given equation: $425 \div 16.95 \times ? = 225$
Approximating $16.95$ to $17$,the equation becomes:
$\frac{425}{17} \times ? = 225$
Calculate $425 \div 17$:
$425 \div 17 = 25$
Now,substitute this back into the equation:
$25 \times ? = 225$
Solve for $?$:
$? = \frac{225}{25}$
$? = 9$