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1: \(6x^2y-9xy^2+3xy\)
\(=3xy\left(2x-3y+1\right)\)
2: \(\left(4-x\right)^2-16\)
\(=\left(4-x-4\right)\left(4-x+4\right)\)
\(=-x\cdot\left(8-x\right)\)
3: \(x^3+9x^2-4x-36\)
\(=x^2\left(x+9\right)-4\left(x+9\right)\)
\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)
1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)
2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)
3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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a, x2 + 2y2 - 3xy + x - 2y = x2 - 4xy + 4y2 + xy - 2y2 + x - 2y
= ( x - 2y )2 + y( x - 2y ) + ( x - 2y )
= ( x - 2y )( x - 2y + y + 1 )
( x - 2y )( x - y + 1 )
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That so easy bài này chệ làm rồi :))))))) mới lớp 7
x4+2x3+3x2+2x+1x4+2x3+3x2+2x+1
Đặt y=x2+1y=x2+1, ta có :
A=x4+2x3+3x2+2x+1A=x4+2x3+3x2+2x+1
=x4+2x2+1+x2+2x+2x3=x4+2x2+1+x2+2x+2x3
=(x2+1)2+2x(x2+1)+x2=(x2+1)2+2x(x2+1)+x2
=y2+2xy+x2=y2+2xy+x2
=(x+y)2=(x2+x+1)
Mấy bài này that so easy
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3xy2 - 6x2y
= 3xy(y - 2x)
\(=3xy\left(y-2x\right)\)