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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
4x^2+2xy-18xy-9y^2
=2x(2x+y)-9y(2x+y)
=(2x-9y)(2x+y)
Sorry bài trên mình làm sai
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
\(=\left(x^2+y^2-5\right)^2-4\left(xy-2\right)^2=\left(x^2+y^2-5+2xy-4\right)\left(x^2+y^2-5-2xy+4\right)\)
\(=\left(\left(x+y\right)^2-9\right)\left(\left(x-y\right)^2-1\right)=\left(x+y-3\right)\left(x+y+3\right)\left(x-y+1\right)\left(x-y-1\right)\)
(x^2+y^2-5)^2 - 4x^2y^2 - 16xy -16
= (x^2 + y^2 - 5)^2 - (4x^2y^2 + 16xy + 16)
= (x^2 + y^2 - 5)^2 - (2xy + 4)^2
= (x^2 + y^2 - 5 - 2xy - 4)(x^2 + y^2 - 5 + 2xy + 4)
= [(x - y)^2 - 9][(x + y)^2 - 1]
=(x-y-3)(x-y+3)(x+y-1)(x+y+1)