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: \(x^2-y^2-2x+2y\)

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

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

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

\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

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

Đặt \(x^2+3x+1=t\)thay vào (1) ta được :

\(\left(t-1\right)\left(t+1\right)+1\)

\(=t^2-1+1\)

\(=t^2\)Thay \(t=x^2+3x+1\)ta được:

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

\(=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+1\right)^2\)

\(=\left[\left(x+\frac{3}{2}\right)-\frac{5}{4}\right]^2\)

\(=\left(x+\frac{3}{2}-\frac{\sqrt{5}}{2}\right)^2\left(x+\frac{3}{2}+\frac{\sqrt{5}}{2}\right)^2\)

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

\(=x^4+x^3+x^2+3x^3+3x^2+3x+x^2+x+1+x^2\)

\(=x^4+4x^3+6x^2+4x+1\)

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

đáp án: a là đúng

A

a) \(5\left(x+2y\right)-15x\left(x+2y\right)=\left(x+2y\right)\left(5-15x\right)\\ =5\left(x+2y\right)\left(1-3x\right)\)

b) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2\\=\left(2x-3\right)^2\)

c) \(\left(3x-2\right)^3-3\left(x-4\right)\left(x+4\right)+\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =27x^3-54x^2+18x-8-3\left(x^2-16\right)+x^3-9x^2+27x-27-\left(x^3+1\right)\\=27x^3-54x^2+18x-8-3x^2+48+x^3-9x^2+27x-27-x^3-1\\ =27x^3-57x^2+36x+12\\ =3\left(3x^3-19x^2+12x+4\right)\)

c) \(27x^3-54x^2+36x-8-3x^2+48+x^3-9x^2+27x-27-x^3-1\\ =27x^3-66x^2+63x+12\\=3\left(9x^3-22x^2+21x+4\right)\)

a, 5( x + 2y )-15x(x + 2y ).

= ( x + 2y ).( 5 -15x )

b, 4x² - 12x + 9 

= ( 2x - 3 )²

\(1,\left(x+2y-3\right)^2-4\left(x+2y-3\right)+4=\left(x+2y-3-2\right)^2=\left(x+2y-5\right)^2\)

\(2,\left(x-y\right)^3-1-3\left(x-y\right)\left(x-y-1\right)=\left(x-y-1\right)\text{[}\left(x-y\right)^2+x-y+1\text{]}-3\left(x-y\right)\left(x-y-1\right)=\left(x-y-1\right)\left(x^2+y^2+x-y+1-3x+3y\right)=\left(x-y-1\right)\left(x^2+y^2-2x+2y+1\right)\)

\(3,\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2=\left(x^2+y^2-17\right)-\left(2xy-8\right)^2=\left(x^2-2xy+y^2-9\right)\left(x^2+y^2+2xy-25\right)=\text{[}\left(x-y\right)^2-3^2\text{]}\text{[}\left(x+y\right)^2-5^2\text{]}=\left(x-y+3\right)\left(x-y-3\right)\left(x+y+5\right)\left(x+y-5\right)\)