a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

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

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

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

`(x^2-x)(x^2-x+1)=6`
Đặt `a=x^2-x(x>=-1/4)`
`pt<=>a(a+1)=6`
`<=>a^2+a-6=0`
`Delta=1+24=25`
`=>a_1=-2(l),a_2=1(tm)`
`<=>x^2-x=1`
`<=>x^2-x-1=0`
`Delta=1+4=5`
`=>x_{12}=(+-sqrt5+1)/2`

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=2\\x^2-x=-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vn\right)\end{matrix}\right.\) 

\(\Rightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)

=>x=-6 hoặc x=1

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=15x^2\)

\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-5x+6\right)-15x^2=0\) (*)

-Đặt \(t=x^2-5x+6\)

(*) \(\Leftrightarrow t\left(t-2x\right)-15x^2=0\)

\(\Leftrightarrow t^2-2xt-15x^2=0\)

\(\Leftrightarrow t^2-5xt+3xt-15x^2=0\)

\(\Leftrightarrow t\left(t-5x\right)+3x\left(t-5x\right)=0\)

\(\Leftrightarrow\left(t-5x\right)\left(t+3x\right)=0\)

\(\Leftrightarrow t-5x=0\) hay \(t+3x=0\)

\(\Leftrightarrow x^2-5x+6-5x=0\) hay \(x^2-5x+6+3x=0\)

\(\Leftrightarrow x^2-10x+6=0\) hay \(x^2-2x+6=0\)

\(\Leftrightarrow x^2-2.5x+25-19=0\) hay \(\left(x-1\right)^2+5=0\) (pt vô nghiệm)

\(\Leftrightarrow\left(x-5\right)^2-19=0\)

\(\Leftrightarrow\left(x-5-\sqrt{19}\right)\left(x-5+\sqrt{19}\right)=0\)

\(\Leftrightarrow x=5+\sqrt{19}\) hay \(x=5-\sqrt{19}\)

-Vậy \(S=\left\{5+\sqrt{19};5-\sqrt{19}\right\}\)

a: =>4x-3x=1-2

=>x=-1

b: =>3x=12

=>x=4

c: =>2(x^2-6)=x(x+3)

=>2x^2-12=x^2+3x

=>x^2-3x-12=0

=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)

Ta có:\(\left|x+5\right|+\left|x-1\right|=\left|x+5\right|+\left|1-x\right|\ge\left|x+5+1-x\right|=6\)

\(\Rightarrow\left|x+5\right|+\left|x+2\right|+\left|x-1\right|=6\Leftrightarrow x=-2\)

* với x ≥ -5

-3x-6=6

⇔ -3x=12

⇔ x=-4 (tm)

*với -5 ≤ x < -2

-x+4=6

⇔ -x=2

⇔ x=-2 (ktm)

* với -2 ≤ x < 1

x+8=6

⇔ x=6-8

⇔ x= -2 (tm)

* với x< 1

3x+4 =6

⇔ 3x=2

⇔ x= \(\dfrac{2}{3}\) (tm)

vậy tập nghiệm của phương trình là S \(\left\{\dfrac{2}{3};-2;-4\right\}\)