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: \(x^2-y^2-2x+2y\)

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

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

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

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

=(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)

\(\left(2x-2013\right)^3+\left(2013-3y\right)^3+\left(3y-2x\right)^3.\)

\(=\left(2x-2013+2013-3y\right)\left[\left(2x-2013\right)^2-\left(2x-2013\right).\left(2013-3y\right)+\left(2013-3y\right)^2\right]+\left(3y-2x\right)^3\)

\(=\left(2x-3y\right).\left(4x^2-8042x+2013^2-4026x+6xy+2013^2-6039y+2013^2-12078y+9y^2\right)+\left(3y-2x\right)^3.\)

\(=\left(2x-3y\right).\left(4x^2+9y^2-12078x-18117y+6xy+3.2013^2\right)-\left(2x-3y\right)^3.\)

\(=\left(2x-3y\right).\left[4x^2+9y^2-12078x-18117y+6xy+3.2013^2-\left(2x-3y\right)^2\right]\)

\(=\left(2x-3y\right).\left(4x^2+9y^2-12078x-18117y+6xy+3.2013^2-4x^2+12xy-9y^2\right).\)

\(=\left(2x-3y\right).\left(-12078x-18117y+18xy+3.2013^2\right).\)

\(=\left(2x-3y\right).9.\left(-1342x-2013y+2xy+1350723\right).\)

\(=9.\left(2x-3y\right).\left[\left(-1342x+2xy\right)+\left(1350723-2013y\right)\right]\)

\(=9.\left(2x-3y\right).\left[-2x\left(671-y\right)+2013\left(671-y\right)\right]\)

\(=9.\left(2x-3y\right).\left(-2x+2013\right)\left(671-y\right)\)

5x(x-1)-3y(x-1)

=5x²-5x-3xy+3y

  5x(x-1)-3y(x-1)

=5x²-5x-3xy+3y

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]

a/Dùng hằng đẳng thức A²-B²=(A+B)(A-B) phân tích được ngay

\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)

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

b/Chắc chỉ phân tích hằng đẳng thức (A-B)²=A²-2ab+B²

\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)

\(=40y^2-395y+748\)

Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa

c/Khai triển biểu thức ban đầu ta được

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

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

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

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

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

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

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

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

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