Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
x3 -2x2 +x- xy2
= x ( x2 - 2x + 1 - y2)
= x\(\left[\left(x-1\right)^2-y^2\right]\)
= x ( x- 1- y) ( x - 1 + y )
X3-2x2+x-xy2
=(x3+x)-(2x2-xy2)
=x(x2+1)-x(2x-y2)
=x(x2-2x+1-y2)
=x[(x-1)2-y]
=x(x-1-y)(x-1+y)
Chúc bạn làm tốt@"
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1+y\right)\left(x-1-y\right)\)
a) \(x^3\) + \(2x^2\) + \(x\)
= \(x\)(\(x^2\) + \(x\) + 1)
= \(x\)(\(x^2+1\))
b)\(x^2\)\(y\) + \(xy\) - \(x\) - \(y\)
= (\(x^2y-y\)) + \(\left(xy-y\right)\)
= \(y\left(x^2-1\right)+\left(xy-y\right)\)
= \(y\left(x-1\right)\left(x+1\right)+y\left(x-1\right)\)
= \(y\left(x-1\right)\left(x+1\right)+1\)
x3 + 2x2y + xy2 - 4x
= x( x2 + 2xy + y2 - 4 )
= x[ ( x + y )2 - 22 ]
= x( x + y - 2 )( x + y + 2 )
\(x^3+2x^2y+xy^2-4x=\left(x^3+x^2y\right)+\left(x^2y+xy^2\right)-4x\)
\(=x^2\left(x+y\right)+xy\left(x+y\right)-4x\)
\(=x\left(x+y\right)^2-4x=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\)
\(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
Ý a có rì đó sai sai nha bn
\(x^2-xy+x^2y-xy^2=x\left(x-y\right)+xy\left(x-y\right)=\left(x-y\right)\left(y+1\right)x\)
\(=x\left(x^2+2x+1-y^2\right)=x\left[\left(x+1\right)^2-y^2\right]=x\left(x+y+1\right)\left(x-y+1\right)\)