[Toán 8] Phân tích thành nhân tử

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phamhuy20011801

$x(y^2-z^2)+y(z^2-x^2)+z(x^2-y^2)\\
=x(y+z)(y-z)+yz^2-yx^2+zx^2-zy^2\\
=x(y+z)(y-z)-(yx^2-zx^2)-(zy^2-yz^2)\\
=x(y+z)(y-z)-x^2(y-z)-yz(y-z)\\
=(xy+xz-x^2-yz)(y-z)\\
=[y(x-z)-x(x-z)](y-z)\\
=(x-y)(y-z)(z-x)$
 
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