Câu hỏi của Dung Thái

Question

Phân tích thành nhân tửa) $\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3$b) $\left(x+y+z\right)\left(xy+yz+zx\right)-xyz$

Cô Hoàng Huyền · Accepted Answer

a) $\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3$$=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]$$=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)$$=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)$Cô nghĩ phân tích đa thức này sẽ đẹp hơn:$\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3$$=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3$$=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3$$=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]$$=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)$$=3\left(x-y\right)\left(y-z\right)\left(z-x\right)$b) $\left(x+y+z\right)\left(xy+yz+zx\right)-xyz$$=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz$$=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz$$=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)$$=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)$$=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]$$=\left(x+y\right)\left(xy+zx+zy+z^2\right)$$=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]$$=\left(x+y\right)\left(y+z\right)\left(z+x\right)$Câu trả lời: a) $\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3$$=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]$$=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)$$=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)$Cô nghĩ phân tích đa thức này sẽ đẹp hơn:$\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3$$=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3$$=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3$$=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]$$=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)$$=3\left(x-y\right)\left(y-z\right)\left(z-x\right)$b) $\left(x+y+z\right)\left(xy+yz+zx\right)-xyz$$=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz$$=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz$$=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)$$=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)$$=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]$$=\left(x+y\right)\left(xy+zx+zy+z^2\right)$$=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]$$=\left(x+y\right)\left(y+z\right)\left(z+x\right)$

Anh Kiên lớp 7 Lê · Answer

a) \left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3(x−y)2+(y−z)3+(z−x)3

=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]=(x−y)2+(y−z+z−x)[(y−z)2−(y−z)(z−x)+(z−x)2]

=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)=(x−y)2+(y−x)(x2+y2+3z2−3yz+xy−3xz)

=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)=(x−y)(x−y−x2−y2−3z2+3yz−xy+3xz

\left(x-y\right)^3+\left(y-z\right)^3+\left

=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3

=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\l

=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\

=\left(x-z\right)\left(

=3\left(x-y\right)\lefb) \left(x+y+z\right)\left(xy+yz+zx\right)-xyzb)(x+y+z)(xy+yz+zx)−xyz

=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz=(xy+yz+zx)(x+y+z)−xyz

=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz=xy(x+y+z)+(yz+zx)(x+y+z)−xyz

=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)Câu trả lời: a) \left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3(x−y)2+(y−z)3+(z−x)3