Câu hỏi của Trần Thanh Tùng

Question

Phân tích đa thức thành nhân tử:$xy\left(x-y\right)-xz\left(x+z\right)+yz\left(2x-y+z\right)$GIÚP MIK VỚI !!!!!!!!!!!!

kudo shinichi · Accepted Answer

$xy\left(x-y\right)-xz\left(x+z\right)+yz\left(2x-y+z\right)$$=xy\left(x-y\right)-xz\left(x+z\right)+yz\left[\left(x-y\right)+\left(x+z\right)\right]$$=xy\left(x-y\right)-xz\left(x+z\right)+yz\left(x-y\right)+yz\left(x+z\right)$$=\left(x-y\right)\left(xy+yz\right)+\left(x+z\right)\left(yz-xz\right)$$=y\left(x-y\right)\left(x+z\right)-z\left(x+z\right)\left(x-y\right)$$=\left(x-y\right)\left(x+z\right)\left(y-z\right)$Câu trả lời: $xy\left(x-y\right)-xz\left(x+z\right)+yz\left(2x-y+z\right)$$=xy\left(x-y\right)-xz\left(x+z\right)+yz\left[\left(x-y\right)+\left(x+z\right)\right]$$=xy\left(x-y\right)-xz\left(x+z\right)+yz\left(x-y\right)+yz\left(x+z\right)$$=\left(x-y\right)\left(xy+yz\right)+\left(x+z\right)\left(yz-xz\right)$$=y\left(x-y\right)\left(x+z\right)-z\left(x+z\right)\left(x-y\right)$$=\left(x-y\right)\left(x+z\right)\left(y-z\right)$

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