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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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x2 - 3x - 40
= x2 - 8x + 5x - 40
= x(x-8) + 5(x-8)
= (x-8)(x+5)
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`#040911`
`x^2 - 3x - 10`
`= x^2 + 2x - 5x - 10`
`= (x^2 + 2x) - (5x + 10)`
`= x(x + 2) - 5(x + 2)`
`= (x - 5)(x + 2)`
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\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(x+y+3\right)\)
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(-x-1)2-(3x-4)2=(x2+2x+1)-(9x2-24x+16)=-8x2+26x-15=\(-8\left(x-\dfrac{5}{2}\right)\left(x-\dfrac{3}{4}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
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x3 - x2 + 3x - 27.
= (x3 - 27) - (x2 - 3x).
= (x - 3) (x2 + 3x + 9) - x (x - 3).
= (x - 3) (x2 + 3x + 9 - x).
= (x - 3) (x2 + 2x + 9).
x3 - x2 + 3x - 27 = ( x3 - 27 ) - ( x2 - 3x )
= ( x - 3 ) ( x2 + 3x + 9 ) - x ( x - 3 )
= ( x - 3 ) ( x2 + 3x + 9 - x )
= ( x - 3 ) ( x2 - 2x + 9 )
* Chúc bạn học tốt
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1: \(\dfrac{A}{B}=\dfrac{2x^4+4x^3-x^3-2x^2-2x^2-4x+x+2}{x+2}\)
\(=2x^3-x^2-2x+1\)
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\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
x2 -3x+2
=x2-2x-x+2
=x(x-2)-(x-2)
=(x-1).(x-2)