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If x2 = y + z, y2 = z + x, z2 = x + y, then what is the value of: \(\frac{1}{x+1}\) + \(\frac{1}{y+1}\) + \(\frac{1}{z+1}\)?(a) 1 (b) 0 (c) –1 (d) 2 |
Answer» (a) 1 x2 = y + z ⇒ x2 + x = x + y + z ⇒ x (x + 1) = x + y + z ⇒ \(\frac{x}{x+y+z}\) = \(\frac{1}{x+1}\) Similarly, \(\frac{1}{y+1}\) = \(\frac{y}{x+y+z}\) and \(\frac{1}{z+1}\) = \(\frac{z}{x+y+z}\) ∴ \(\frac{1}{x+1}\) + \(\frac{1}{y+1}\) + \(\frac{1}{z+1}\) = \(\frac{x}{x+y+z}\) + \(\frac{y}{x+y+z}\) + \(\frac{z}{x+y+z}\) = \(\frac{x+y+z}{x+y+z}\) = 1. |
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