Question

If $X^2=Y+Z$,$Y^2=Z+X$,$Z^2=X+Y$, than the value of $\frac{1}{X+1}+\frac{1}{Y+1}+\frac{1}{Z+1}$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is B $1$

$x^2=y+z$ $y^2=z+x$ $z^2=x+y$

$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$

$=1-\frac{x}{1+x}+1-\frac{y}{1+xy}+1-\frac{z}{1+z}$

$=1-\frac{x}{1+x}+1-\frac{y}{1+y}+1-\frac{z}{1+z}$

$=3-(\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z})$

$=3-(\frac{x^2}{x+x^2}+\frac{y^2}{y+y^2}+\frac{z^2}{z+z^2})$

$=3-(\frac{y+z}{x+y+z}+\frac{x+z}{x+y+z}+\frac{x+y}{x+y+z})$

$=3-2\left(\frac{x+y+z}{x+y+z}\right)$

$=3-2$

$=1$

B is correct