d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)

=>\(x+3-6\left(x-2\right)=-5\)

=>x+3-6x+12=-5

=>-5x+15=-5

=>-5x=-20

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)

=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)

=>\(2x^2-4x+8-2x^2-16=5x+10\)

=>5x+10=-4x-8

=>9x=-18

=>x=-2(loại)

f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)

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

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

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

=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)

=>\(\left(x^2-1\right)\cdot\left(x^3+1-x^3+1\right)=2\left(x^2+4x+4\right)\)

=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)

=>8x=-10

=>x=-5/4(nhận)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

a: Ta có: \(6-4x=5(x+3)+3\)

\(\Leftrightarrow6-4x-5x-12-3=0\)

\(\Leftrightarrow-9x=9\)

hay x=-1

b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)

\(\Leftrightarrow15x+45-30=10x-30+5x+25\)

\(\Leftrightarrow15=-5\left(loại\right)\)

c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)

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

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

\(\Leftrightarrow2+x-2=x^2+2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

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

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

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

Suy ra: \(x^2+4x+4+2x-4=x^2\)

\(\Leftrightarrow6x=0\)

hay \(x=0\left(nhận\right)\)

2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)

Suy ra: \(x+6-2x+12=3x+6\)

\(\Leftrightarrow-x-3x=6-18=-12\)

hay \(x=3\left(nhận\right)\)

Lời giải:
1. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)

\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)

\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)

2. ĐKXĐ: $x\neq \pm 6$

PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)

\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)

\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)

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

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

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

\(\Leftrightarrow\dfrac{x^2+2x2+2^2+2x-4-x^2}{\left(x-2\right)\left(x+2\right)}\)=0

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

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

\(\Leftrightarrow6x=0\)

\(\Rightarrow x=0\)

2) \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{2}{x+6}-\dfrac{\left(3x+6\right)}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{1\left(x+6\right)-2\left(x-6\right)-\left(3x+6\right)}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{x+6-2x+12-3x-6}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{x-2x-3x+6-6+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow\dfrac{-4x+12}{\left(x-6\right)\left(x+6\right)}\)=0

\(\Leftrightarrow-4x+12=0\)

\(\Leftrightarrow-4x=12\)

\(\Rightarrow x=3\)

a, đk : x khác 5;-6 

\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm) 

b, đk : x khác 1;3 

\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)

pt vô nghiệm 

a, đk : x khác 5;-6 

(tm) 

b, đk : x khác 1;3 

x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)

pt vô nghiệm 

a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>x=0(nhận)

b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=-6

hay x=-3(nhận)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)

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

\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)

\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)

\(< =>3x=6\\ < =>x=2\left(tm\right)\)

ĐKXĐ: \(x\ne\left\{-2;3\right\}\)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

\(\Leftrightarrow\dfrac{9x-2}{\left(x+2\right)\left(x-3\right)}+\dfrac{2x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)

\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\left(loại\right)\)

Vậy: PT vô nghiệm.

d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

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

ĐKXĐ: \(x\ne3;x\ne-2\)

\(\left(1\right)\Leftrightarrow x\left(x+2\right)+x\left(x-3\right)=3x+6\)

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

\(\Leftrightarrow2x^2-4x-6=0\)

\(\Leftrightarrow2\left(x^2-2x-3\right)=0\)

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

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

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0;x-3=0\)

*) \(x+1=0\)

\(\Leftrightarrow x=-1\) (nhận)

*) \(x-3=0\)

\(\Leftrightarrow x=3\) (loại)

Vậy \(S=\left\{-1\right\}\)