Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
Option 1: $0$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $3$
Correct Answer: $\frac{3}{2}$
Solution : Given: $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$ ⇒ $\frac{1}{x+2}=\frac{1}{3}$ $\therefore x=1$ Similar way, $y=6, z=2662$ Now, $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$ = $\frac{1}{1+1}+\frac{6}{6+6}+\frac{2662}{2662+2662}$ = $\frac{1}{2}+\frac{6}{12}+\frac{2662}{5324}$ = $\frac{7986}{5324}$ = $\frac{3}{2}$ Hence, the correct answer is $\frac{3}{2}$.
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Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:
Option 1: –1
Option 2: 1
Option 3: 2
Option 4: 4
Question : If $x+y=2z$, then the value of $\frac{x}{x-z}+\frac{z}{y-z}$ is:
Option 1: $1$
Option 2: $3$
Option 3: $\frac{1}{2}$
Option 4: $2$
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}$, where $x \neq y \neq z$, is:
Option 2: $\frac{1}{(x+y+z)}$
Option 3: $\frac{1}{(x+y)(y+z)(z+x)}$
Option 4: $1$
Question : If $\frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$, what is the value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$
Option 1: –3
Option 2: 0
Option 3: 15
Option 4: 21
Question : If $x+y+z=17, x y z=171$ and $x y+y z+z x=111$, then the value of $\sqrt[3]{\left(x^3+y^3+z^3+x y z\right)}$ is:
Option 1: –64
Option 2: 4
Option 3: 0
Option 4: –4
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