1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

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

1) Ta có: \(x^3+2x^2-6x-27\)

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

\(2x-1^3+8\)

\(=2x-9\)

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

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

_________

\(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

_______________

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

\(=8x^3-12x^2+6x-1-1\)

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

\(=x^3+8x^3-12x^2+6x-1\)

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

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

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

b: 8x^3-12x^2+6x-1

=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

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

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

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

1) \(\frac{3}{x^2-4y^2}\)

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

Phân thức xác định khi \(\left(x-2y\right)\left(x+2y\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x-2y\ne0\\x+2y\ne0\end{cases}}\Rightarrow x\ne\pm2y\)

2) \(\frac{2x}{8x^3+12x^2+6x+1}\)

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

Phân thức xác định khi \(\left(2x+1\right)^3\ne0\)

\(\Rightarrow2x+1\ne0\)

\(\Rightarrow x\ne-\frac{1}{2}\)

3) \(\frac{5}{2x-3x^2}\)

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

Phân thức xác định khi : \(x\left(2-3x\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x\ne0\\2-3x\ne0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ne0\\x\ne\frac{2}{3}\end{cases}}\)

Bài 1:tìm x ,biết:

a) (2x - 1)(3x + 2) - 6x(x + 1) = 0

\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)

\(\Leftrightarrow-5x=2\)

\(\Leftrightarrow x=\frac{-2}{5}\)

b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)

\(\Leftrightarrow-10x=-4\)

\(\Leftrightarrow x=\frac{2}{5}\)

c) \(4x^2-1=2\left(2x+1\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)

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

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

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

c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)

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

d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)

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

e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)

\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)

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

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

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

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

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

Bài 2

a) 4x(x-3)-3x+9

=4x(x-3)-3(x-3)

= (x-3)(4x-3)

b) x³+2x²-2x-4

=(x³+2x²)-(2x+4)

=x²(x+2)-2(x+2)

=(x+2)(x²-2)

c) 4x²-4y+4y-1

=4x²-1

=(2x-1)(2x+1)

d) x⁵-x

=x(x⁴-1)

=x(x²-1)(x²+1)

a) 4x(x-3)-3x+9

= 4x(x-3) - 3(x-3)

= (x-3)(4x-3)

b)x³ + 2x² - 2x - 4

= x²(x + 2) - 2(x + 2)

= (x+2)(x²-2)

c) 4x² - 4y +4y -1

= [(2x)²-1²] + (-4y+4y)

= (2x+1)(2x-1)

d) x⁵-x

= x(x⁴ - 1)

a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4 
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))

a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)

b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)

d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)