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If x + y + z = 0, then \(\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}\) =

(a) 4 

(b) 2 

(c) 3 

(d) 1

1 Answer

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Best answer

(d) 1

x + y + z = 0 ⇒ x = – y – z                           ...(i) 

y = – x – z                                                   ...(ii) 

z = – x – y                                                 ...(iii)

Now, \(\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}\)

\(\frac{x^2}{x^2+x.x+yz}+\frac{y^2}{y^2+y,y +zx}+\frac{z^2}{z^2+z.z+xy}\)

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