[Toán 8] Rút gọn phân thức

H

hiensau99

$(y-z)^2+(z-x)^2+(x-y)^2= 2(x^2+y^2+z^2)-2xy-2xz-2yz= 3(x^2+y^2+z^2)-(x^2+y^2+z^2+2xy+2xz+2yz)=3(x^2+y^2+z^2)-(x+y+z)^2= 3(x^2+y^2+z^2)$

Vậy: $\frac{x^2+y^2+z^2}{(y-z)^2+(z-x)^2+(x-y)^2} = \frac{1}{3}$
 
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