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)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

ĐKXĐ

x≠3 ; x≠-3

ĐKXĐ x≠3 ; x≠-3

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

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

⇔2x²-6x-x+3=2x²+6x+x+3

⇔2x²-2x²-7x-6x=3-3

⇔ -13x=0

⇔x=0 (tm)

vậy phương trình trên có tập n^o S={0}

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}

Pt trên có MSC là \(\left(x-1\right)\left(x^2+x+1\right)\)

Quy đồng mẫu số :

\(\dfrac{1}{x-1}+\dfrac{7x-10}{x^3-1}-\dfrac{3}{x^2+x+1}=0\)

( ĐKXĐ \(x\ne1\))

\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{x^3-1}-\dfrac{3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+x+1+7x-10-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{x^2+5x-6}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=-6\left(TMĐK\right)\end{matrix}\right.\)

Vậy \(S=\left\{-6\right\}\)

ĐKXĐ: \(x\ne1\); \(x\ne-1\)

\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Rightarrow x^2+x+1+7x-10-3x+3=0\)

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

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

\(\Leftrightarrow x-1=0\) ; \(x+6=0\)

+) \(x-1=0\)

\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)

+) \(x+6=0\)

\(\Leftrightarrow x=-6\) (Thỏa mãn ĐKXĐ)

Tập nghiệm: \(S=\left\{-6\right\}\)

với x>0 thì pt luôn xác định.

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

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

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

\(x\left(x^2+x-2\right)=0\Rightarrow x=0\) hoặc \(x^2+x-2=0\)

x=0 hoac (x\(^2\)-1) +(x-1) =0

x=0 hoặc (x-1)(x+2)=0

x=0 hoax x=1 hoặc x=2 vỉ x>0 nên pt có 2 nghiệm là x=1 , x=2.

x = 0

X=1

\(\Leftrightarrow x^2-x-x-1=-1\)

=>x(x-2)=0

=>x=2

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)

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