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Simplify \(\frac{x^2-(y-2z)^2}{x-y+2z}\) + \(\frac{y^2-(2x-z)^2}{y+2x-z}\) + \(\frac{z^2-(x-2y)^2}{z-x+2y}\) is...............A) 1 B) x + y + z C) 0 D) x – y – z |
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Answer» Correct option is (C) 0 \(\frac{x^2-(y-2z)^2}{x-y+2z}+\frac{y^2-(2x-z)^2}{y+2x-z}+\frac{z^2-(x-2y)^2}{z-x+2y}\) \(=\frac{(x-(y-2z))(x+y-2z)}{x-y+2z}+\frac{(y-(2x-z))(y+2x-z)}{y+2x-z}+\frac{(z-(x-2y))(z+x-2y)}{z-x+2y}\) \((\because(a+b)(a-b)=a^2-b^2)\) \(=\frac{(x-y+2z)(x+y-2z)}{x-y+2z}+\frac{(y-2x+z)(y+2x-z)}{y+2x-z}+\frac{(z-x+2y)(z+x-2y)}{z-x+2y}\) = (x+y-2z) + (y-2x+z) + (z+x-2y) = (x+x-2x) + (y+y-2y) + (-2z+z+z) = 0+0+0 = 0 Correct option is C) 0 |
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