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

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)

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

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

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

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

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

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

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

b) Bạn có thể viết kiểu latex được không ạ ?

a: \(\Leftrightarrow\left(-x+3\right)\left(x+6\right)=18\)

\(\Leftrightarrow-x^2-6x+3x+18-18=0\)

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

=>x=0 hoặc x=-3

b: \(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{\sqrt{21}}{3}-1;\dfrac{-\sqrt{21}}{3}-1\right\}\)

c: =>x(3x-5)=0

=>x=0 hoặc x=5/3

d: =>(x-2)(x+2)=0

=>x=2 hoặc x=-2

a. 

$x^4-6x^2+9=0$

$\Leftrightarrow (x^2-3)^2=0$

$\Leftrightarrow x^2-3=0$

$\Leftrightarrow x^2=3$

$\Leftrightarrow x=\pm \sqrt{3}$

b.

$8x^3+12x^2+6x-63=0$

$\Leftrightarrow (8x^2+12x^2+6x+1)-64=0$

$\Leftrightarrow (2x+1)^3=64=4^3$

$\Leftrightarrow 2x+1=4$

$\Leftrightarrow x=\frac{3}{2}$

c. $(3-2x)^2-25=0$

$\Leftrightarrow (3-2x)^2-5^2=0$

$\Leftrightarrow (3-2x-5)(3-2x+5)=0$

$\Leftrightarrow (-2-2x)(8-2x)=0$

$\Leftrightarrow -2-2x=0$ hoặc $8-2x=0$

$\Leftrightarrow x=-1$ hoặc $x=4$

d.

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

$\Leftrightarrow (x+1)^2[6-2(x+1)]+2(x^3-1)=1$

$\Leftrightarrow (x+1)^2(4-2x)+2x^3-3=0$

$\Leftrightarrow 6x+1=0$

$\Leftrightarrow x=\frac{-1}{6}$

e. $(x-2)^2-(x-2)(x+2)=0$

$\Leftrightarrow (x-2)[(x-2)-(x+2)]=0$

$\Leftrightarrow (x-2)(-4)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

f. $x^2-4x+4=25$

$\Leftrightarrow (x-2)^2=5^2=(-5)^2$

$\Leftrightarrow x-2=5$ hoặc $x-2=-5$

$\Leftrightarrow x=7$ hoặc $x=-3$

a. 

${}^{}-6{}^{}+9=0$

$\iff ({}^{}-3{}^{}=0$

$\iff {}^{}-3=0$

$\iff {}^{}=3$

$\iff x=\pm \sqrt{3}$

b.

$8{}^{}+12{}^{}+6x-63=0$

$\iff (8{}^{}+12{}^{}+6x+1)-64=0$

$\iff (2x+1{}^{}=64={}^{}$

$\iff 2x+1$

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)

1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)

2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)

3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)

1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
 

5: \(4\left(x-3\right)^2+2x\left(3-x\right)\)

\(=\left(x-3\right)\left(4x-12\right)-2x\left(x-3\right)\)

\(=\left(x-3\right)\left(2x-12\right)\)

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

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

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

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

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

1) Ta có: \(x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

\(\left(\dfrac{1}{2}x+3\right)\left(2x^3-6x^2+4x+10\right)\\ =\dfrac{1}{2}x.2x^3-\dfrac{1}{2}x.6x^2+\dfrac{1}{2}x.4x+\dfrac{1}{2}x.10+3.2x^3-3.6x^2+3.4x+3.10\\ =x^4-3x^3+2x^2+5x-6x^3-18x^2+12x+30\\ =x^4-9x^3+20x^2+17x+30\)