If x2 = y + z, y2 = z + x, z2 = x + y, then the value of $$\frac{1}{{x + 1}}$$ + $$\frac{1}{{y + 1}}$$ + $$\frac{1}{{z + 1}} = \,?$$
A. -1
B. 1
C. 2
D. -2
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {x^2} = y + z \cr & {y^2} = z + x \cr & {z^2} = x + y \cr & \Rightarrow {x^2} + x = x + y + z \cr & {\text{Adding }}x{\text{ on both sides }} \cr & x\left( {x + 1} \right) = x + y + z \cr & \frac{1}{{\left( {x + 1} \right)}} = \frac{x}{{x + y + z}} \cr & {\text{Similarly,}} \cr & \frac{1}{{\left( {y + 1} \right)}} = \frac{y}{{x + y + z}} \cr & \frac{1}{{\left( {z + 1} \right)}} = \frac{z}{{x + y + z}} \cr & \therefore \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}} \cr & = \frac{x}{{x + y + z}} + \frac{y}{{x + y + z}} + \frac{z}{{x + y + z}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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