[Toán 8]Rút gọn phân thức

C

chuotdelux

x + y + z = 0

\Rightarrow x + y = - z

\Rightarrow $(x + y)^2 = z^2$

\Rightarrow $x^2 + y^2 + 2xy = z^2$

\Rightarrow $x^2 + y^2 - z^2 = - 2xy$

\Rightarrow $- xy = \dfrac{x^2 + y^2 - z^2}{2}$

Tương tự có : $- yz = \dfrac{y^2 + z^2 - x^2}{2}; - xz = \dfrac{x^2 + z^2 - y^2}{2}$

$\rightarrow - xy - yz - xz = \dfrac{x^2 + y^2 + z^2}{2}$

Ta có : $\dfrac{x^2 + y^2 + z^2}{(y - z)^2 + (z - x)^2 + ( x - y)^2} = \dfrac{x^2 + y^2 + z^2}{2(x^2 + y^2 + z^2 - yz - xy - xz)} = \dfrac{x^2 + y^2 + z^2}{2(x^2 + y^2 + z^2 + \dfrac{x^2 + y^2 + z^2}{2}} = \dfrac{x^2 + y^2 + z^2}{3(x^2 + y^2 + z^2} = \dfrac{1}{3}$
 
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