A) \[-1\]
B) 1
C) 2
D) 4
Correct Answer: B
Solution :
\[{{x}^{2}}=y+z\] \[\Rightarrow \] \[{{x}^{2}}+x=x+y+z\] \[\Rightarrow \] \[x(x+1)=x+y+z\] ? (i) Similarly, \[y(y+1)=x+y+z\] ? (ii) and \[z(z+1)=x+y+z\] ?(iii) \[\therefore \] \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\] \[=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\] \[=\frac{x+y+z}{x+y+z}=1\]You need to login to perform this action.
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