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

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

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

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

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

Chúc bạn học tốt ~ 

Ta có : (x³ + 3x + 1)(x³ + 3x + 2) - 6

= (x³ + 3x + 1,5 - 0,5)(x³ + 3x + 1,5 + 0,5) - 6

= (x³ + 3x + 1,5)² - 0,5² - 6

= (x³ + 3x + 1,5)² - 6,25

= (x³ + 3x + 1,5 - 2,5) (x³ + 3x + 1,5 + 2,5)

= (x³ + 3x - 1) (x³ + 3x + 3)

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: \(x^2-y^2-2x+2y\)

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

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

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

\(A=4x^2+6x=2x\left(2x+3\right)\)

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

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

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

B

c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)

Đặt \(x^2+3x+1,5=a\)

\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)

\(\Rightarrow A=a^2-0,25-6\)

\(\Rightarrow A=a^2-\frac{25}{4}\)

\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)

Thay \(a=x^2+3x+0,5\)vào A ta có :

\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)

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

c, Đặt \(x^2+3x+2=a\)

Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)

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

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

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

Câu d làm tương tự .

Gợi ý : (x+3)(x+5) = x² + 8x + 15 

đặt bằng a rồi giải tiếp

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 : \(x^3-3x^2y+3xy^2-y^3+y-x\)

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

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

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

phân tích đa thức thành nhân tử cơ mà 

=(x-y)³-(x-y)

=(x-y)[(x-y)²-1]

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