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![](https://rs.olm.vn/images/avt/0.png?1311)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)
\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
a, = 2x.(x-2)
b, = (x^2+y^2+2xy)-(2x+2y)
= (x+y)^2-2.(x+y)
= (x+y).(x+y-2)
2,
a,<=> x^2-1-x^2-2x = 3
<=> -2x-1=3
<=> -2x=4
<=> x=4 : (-2) = -2
b, <=>(x^2-4x+4)-7=0
<=>(x-2)^2-7=0
<=> (x-2)^2=7
=> x-2=+-\(\sqrt{7}\)
<=> x=2+-\(\sqrt{7}\)
k mk nha
a, \(2x-4x\)
\(=-2x\)
b, \(x^2+y^2+2xy-2x-2y\)
\(=\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
a, \(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow x^2-1-x^2-2x=3\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\)
b,\(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
a: 2x+4=2(x+2)
b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)