bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

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

\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

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

\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)

\(=\dfrac{1}{6\left(x^2+x+1\right)}\)

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

\(=x\left(9x^2-y^2\right)\)

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

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

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

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

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

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

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

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

\(=\left(6-x+3y\right)\left(6+x-3y\right)\)

5: \(x^4-6x^2+5\)

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

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

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

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

6: \(9x^2-6x-y^2+2y\)

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

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

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

Thiếu đề đó

đúng đề rồi

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

=>\(-\left(x-2\right)^3-6x\left(x-2\right)=0\)

=>\(\left(x-2\right)^3+6x\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x^2-4x+4+6x\right)=0\)

=>\(\left(x-2\right)\left(x^2+2x+4\right)=0\)

=>x-2=0

=>x=2

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

6 - 2x - 3x -3 + 6x = 2

3 + x = 2

x  = 2 - 3 

x = -1

vậy số nguyên x là -1 

\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\\ \Leftrightarrow12x-12=0\\ \Leftrightarrow12x=12\\ \Leftrightarrow x=1\)

\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\\ \Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\\ \Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\\ \Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\\ \Leftrightarrow6x^2-6x^2+5x+6=1\\ \Leftrightarrow5x=-5\\ \Leftrightarrow x=-1\)

x³- 6x³ -x + 30 = 0  

x³ + 2x²- 8x²- 16x + 15x + 30 = 0  

x² ( x + 2 ) - 8x ( x + 2 ) + 15 ( x + 2 ) = 0  

( x + 2 )( x² - 8x + 15 ) = 0  

x + 2 = 0 hoặc x² - 8x + 15 = 0

 x = - 2 hoặc ( x - 4 )² - 1 = 0  

x = - 2 hoặc ( x - 4 - 1 ) ( x - 4 + 1 ) = 0

 x = - 2 hoặcx = 5 hoặc x = 3

yeah               

Ta có: \(x^3+6x^2+9x=0\)

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

hay \(x\in\left\{0;-3\right\}\)

x³+6x²+9x=0

⇒x(x²+6x+9)=0

⇒x(x+3)²=0

⇒\(\left[{}\begin{matrix}x=0\\\left(x+3\right)^2=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

\(pt< =>x\left(x^2+6x+9\right)=0< =>x\left(x+3\right)^2=0\)

\(=>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

6(x-3)(x-4)-6x(x-2)=4

<=>(6x-18)(x-4)-6x²+12=4

<=>6x²-24x-18x+72-6x²+12=4

<=>-30x+72=4

<=>-30x=-68

<=>x=34/15