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

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

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

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

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

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

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

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

=   (x + 2y - 3)²  - 2. 2. (x + 2y - 3)  +  2²   (hằng đẳng thức số 2, bình phương  của một hiệu)

=  ( x + 2y - 3 - 2)²

= ( x + 2y - 5)²

Câu a :

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

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

\(=4x^2-19x-5\)

Câu b :

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

\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)

\(=9x^2-24x+2\)

(1-x^2)^2-4x(1-x^2)

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

=(1-x^2)(1-x)^2

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

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

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

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: \(x^2-y^2-2x+2y\)

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

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

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

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

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

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

\(=\)\(\left(3x-2\right)\left(3x-6\right)\)

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

Chúc bạn học tốt ~ 

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)

\(=9x^2+18x+9-9x^2+12x-4\)

\(=30x+5\)

\(=5\left(6x+1\right)\)

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

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

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

\(=5\left(6x+1\right)\)

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

=\(\left(9x^2+18x+9\right)-\left(9x^2-12x+4\right)\)

=\(9x^2+18x+9-9x^2+12x-4\)

=\(5\left(6x+1\right)\)

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