Question

rút gọn
x^2 + y^2 + z^2 / (y-z)^2 + (z-x)^2 + (x-y)^2 biết x+ y + z = 0

Answer 1

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Answer 2

Đặt A =\(\frac{x^2+y^2+z^2}{(y-z)^2 + (z-x)^2 + ( x-y)^2}\) (ĐKXĐ : \(x\ne y \ne z\))

=> A = \(\frac{x^2+y^2+z^2+2xy+2xz+2yz - 2xy-2yz-2xz}{y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2}\)

= \(\frac{(x+y+x)^2-2(xy+xz+yz)}{2(x^2+y^2+z^2)- 2(xy+xz+yz)}\)

= \(\frac{-2(xy+xz+yz)}{2[(x+y+z)^2-2(xy+xz+yz)] - 2(xy+xz+yz)}\) Vì \(x+y+z=0\)

= \(\frac{-2(xy+xz+yz)}{2[-2(xy+xz+yz)]-2(xy+xz+yz)}\) Vì \(x+y+z=0\)

= \(\frac{-2(xy+xz+yz)}{-4(xy+xz+yz)-2(xy+xz+yz)}\)

= \(\frac{-2(xy+xz+yz)}{-6(xy+xz+yz)}\)

= \(\frac{1}{3}\)