Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Option 1: $\frac{11}{13}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{16}{9}$
Option 4: $\frac{9}{16}$
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Correct Answer: $\frac{13}{11}$
Solution : Given: $(x-\frac{1}{3})^2+(y-4)^2=0$ $(x-\frac{1}{3})^2=0$ , $(y-4)^2=0$ So, $x=\frac{1}{3}, y=4$ Now, $\frac{y+x}{y-x}=\frac{4+\frac{1}{3}}{4-\frac{1}{3}}=\frac{13}{11}$ Hence, the correct answer is $\frac{13}{11}$.
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Question : If $(4 y-\frac{4}{y})=13$, find the value of $(y^2+\frac{1}{y^2})$.
Option 1: $12 \frac{11}{16}$
Option 2: $10 \frac{9}{16}$
Option 3: $12 \frac{9}{16}$
Option 4: $8 \frac{9}{16}$
Question : If $\frac{11-13x}{x}+\frac{11-13y}{y}+\frac{11-13z}{z}=5$, then what is the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$?
Option 1: $1$
Option 3: $\frac{13}{5}$
Option 4: $4$
Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Option 1: $\frac{3}{8}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{4}{9}$
Option 4: $\frac{11}{13}$
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Option 1: 1
Option 2: 0
Option 3: 2
Option 4: 3
Question : If $2 x-y=2$ and $x y=\frac{3}{2}$, then what is the value of $x^3-\frac{y^3}{8}?$
Option 1: $\frac{9}{2}$
Option 2: $-\frac{5}{4}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{13}{4}$
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