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

• A. $-1$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$1$

$x^2=y+z,y^2=x+z,z^2=x+y⟶1\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{1+x-x}{1+x}+\frac{1+y-y}{1+y}+\frac{1+z-z}{1+z}=1-\frac{x}{1+x}+1\frac{y}{1+y}+1-\frac{z}{1+z}=3-(\frac{x^2}{x+x^2}+\frac{y^2}{y+y^2}+\frac{z^2}{z+z^2})$

$($ Multiplied numerator and denominator byx,y and z in the 3 terms $)$

$=3-(\frac{y+z}{x+y+z}+\frac{z+x}{x+y+z}+\frac{x+y}{x+y+z})=3-2\frac{(x+y+z)}{(x+y+z)}=3-2=1$