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)}

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

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)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

7,x²-2xy+y²+3x-3y=(x-y)²+3(x-y)=(x-y)(x-y+3)

8,x⁴+4=(x⁴+4x²+4)-4x²=(x²+2)²-(2x)²=(x²-2x+2)(x²+2x+2)

9,4x(x+1)²-5x²(x+1)-4.(x+1)=(x+1)\(\left[4x\left(x+1\right)-5x^2-4\right]\)=(x+1)(4x²+4x-5x²-4)=(x+1)(-x²+4x-4)=-(x+1)(x-2)²

7: \(x^2-2xy+y^2+3x-3y\)

\(=\left(x-y\right)^2+3\cdot\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y+3\right)\)

8: \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

9: \(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right)\)

\(=\left(x+1\right)\left(4x^2+4x-5x^2-4\right)\)

\(=\left(x+1\right)\left(-x^2+4x-4\right)\)

\(=-\left(x+1\right)\left(x-2\right)^2\)

10: \(\left(1+2x\right)\left(1-2x\right)-\left(x+2\right)\left(x-2\right)\)

\(=1-4x^2-x^2+4\)

\(=-5x^2+5\)

\(=-5\left(x-1\right)\left(x+1\right)\)

\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)

\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)

\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)

\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)

a) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

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,\(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

1)

a)\(2x-4x=x\left(2-4\right)=-2x\)

b)\(x^2+y^2+2xy-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x+y-2\right)\)

2)

a)\(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)

\(\Leftrightarrow\left(x^2-1\right)-x\left(x+2\right)=3\)

\(\Leftrightarrow\left(x^2-1\right)-x^2-2x=3\)

\(\Leftrightarrow x^2-1-x^2-2x=3\)

\(\Leftrightarrow-1-2x=3\)

\(\Leftrightarrow-2x=4\)

\(\Leftrightarrow=-2\)

b)\(x^2-4x+3=0\)

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

P/ s tham khảo nha