Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

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

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

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

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

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

x³+y(1-3x²)+x(3y²-1)-y³

= x³-3x²y+3xy²-y³+y-x

=(x-y)³ -(x-y)

=(x-y)(x²-2xy+y²-1)

cái chỗ kia giải thích dùm mìh đy : \(x^3-3x^2y+3xy^2-y^3+y-x\)

bạn nhân ra ra được mà chẳng hạn x(x+1)=x²+x

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

\(=x^3-3x^2y+3xy^2-y^3+y-x\)

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

phân tích đa thức thành nhân tử cơ mà 

=(x-y)³-(x-y)

=(x-y)[(x-y)²-1]

\(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)

=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)²

=(2x-3y)(x-y-2y+3x)-(4x-3y)²

=(2x-3y)(4x-3y)-(4x-3y)²

^{=(4x-3y)(2x-3y-4x+3y)}

=(4x-3y))(-2x)

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

`a)2x(x-3)-x+3=0`

`<=>2x(x-3)-(x-3)=0`

`<=>(x-3)(2x-1)=0`

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

`<=>(x+2)(x+3)=0`

`<=>[(x=-2),(x=-3):}`

a) \(A=x^2-2xy+y^2+3x-3y-4\)

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

\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)

\(=\left(x-y-1\right)\left(x-y+1+3\right)\)

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

x² + y² - 3x - 3y + 2xy

= ( x² + 2xy + y² ) - ( 3x + 3y )

= ( x + y )² - 3( x + y )

= ( x + y )( x + y - 3 )

b) ( x² - 4x )² - 2( x - 2 )² - 7 

= ( x² - 4x )² - 2( x² - 4x + 4 ) - 7 (*)

Đặt t = x² - 4x

(*) <=> t² - 2( t + 4 ) - 7

       = t² - 2t - 8 - 7

       = t² - 2t - 15

       = t² + 3t - 5t - 15

       = t( t + 3 ) - 5( t + 3 )

       = ( t + 3 )( t - 5 )

       = ( x² - 4x + 3 )( x² - 4x - 5 ) 

       = ( x² - x - 3x + 3 )( x² + x - 5x - 5 )

       = [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 ) ]

       = ( x - 1 )( x - 3 )( x + 1 )( x - 5 )

a) Ta có: \(x^2+y^2-3x-3y+2xy\)

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

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

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

        \(=\left(x+y-1\right)^2-\left(\sqrt{x+y+1}\right)^2\)

        \(=\left(x+y-1+\sqrt{x+y+1}\right)\left(x+y-1-\sqrt{x+y+1}\right)\)

\(3x^2-3y^2-2\left(x-y\right)^2\)

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

\(=3x^2-3y^2-2x^2+4xy-2y^2\)

\(=x^2+4xy-5y^2\)

\(=x^2+4xy+4y^2-9y^2\)

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

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

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

a: \(x^2-2xy+y^2+3x-3y-4\)

\(=\left(x-y\right)^2+3\left(x-y\right)-4\)

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