Sửa lại đề nha:

 \(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)

⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)

⇔ \(x^2+8x+12=32\)

⇔ \(x^2+8x-20=0\)

⇔ \(\left(x-2\right)\left(x+10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

Ta có : 1+\(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}\) (đkxđ x\(\ne\pm2\) )

\(\Leftrightarrow\) \(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}-1\)

\(\Leftrightarrow\)\(\dfrac{1}{x+2}=\dfrac{12-\left(8-x^3\right)}{8-x^3}\)

\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{x^3+4}{8-x^3}\)

\(\Leftrightarrow8-x^3=\left(x+2\right)\left(x^3+4\right)\)

\(\Leftrightarrow8-x^3=x^4+4x+2x^3+8\)

\(\Leftrightarrow-x^3-x^4-4x-2x^3=8-8\)

\(\Leftrightarrow-x^4-3x^3-4x=0\)

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

\(\Rightarrow-x=0\)\(\Rightarrow x=0\) (TM x\(\ne\pm2\))

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}

`1/x+1/(x+2)=5/12`

ĐK:`x ne 0,x ne -2`

`<=>(x+2+x)/(x^2+2x)=5/12`

`<=>(2x+2)/(x^2+2x)=5/12`

`<=>24x+24=5x^2+10x`

`<=>5x^2-14x-24=0`

Ta có:`Delta'=49+24.5`

`=49+120=169`

`=>x_1=-6/5,x_2=4`

Vậy `S={4,-6/5}`

$ĐKXĐ : x \neq 0, x \neq -2$

Ta có : $\dfrac{1}{x} + \dfrac{1}{x+2} = \dfrac{5}{12}$

$\to \dfrac{2x+2}{x.(x+2)} = \dfrac{5}{12}$

$\to (2x+2).12 = x.(x+2).5$

$\to 24x + 24 = 5x^2 + 10x$

$\to 5x^2 - 14x - 24 = 0 $

$\to (x-4).(5x+6) = 0 $

S\to$ \(\left[{}\begin{matrix}x=4\\x=-\dfrac{6}{5}\end{matrix}\right.\)  ( thỏa mãn ĐKXĐ )

Vậy :....

b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)

\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)

Bình phương 2 vế rút gọn

\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)

Đặt \(\sqrt{x^4-x^2}=a\)

\(\Rightarrow a^2-4\sqrt{3}a+12=0\)

\(\Leftrightarrow a=2\sqrt{3}\)

\(\Leftrightarrow x^4-x^2=12\)

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

Câu a xem lại đề đúng không b. Do nghiệm xấu lắm

a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)

=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)

=>\(x^2+2x+1-x^2+2x-1=16\)

=>4x=16=>x=4

b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)

=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

=>\(12-x^2-3x-2+x^2+5x-14=0\)

=>2x-4=0=>2x=4=>x=2

c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)

=>\(12=x^3+8+x^2-2x+4\)

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

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

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

=>\(x\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

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

=>x=?

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\)

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

\(\Leftrightarrow\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm3;x\ne-1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)+8\left(x+3\right)=2x\left(x+3\right)\)

\(\Leftrightarrow x^2-2x+1+8x+24=2x^2+6x\)

\(\Leftrightarrow-x^2+25=0\)

\(\Leftrightarrow x^2-25=0\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy \(S=\left\{-5;5\right\}\)