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:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

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

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

\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)

\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)

\(\Leftrightarrow9x^2-144x-324=0\)

\(\Leftrightarrow x^2-16x-36=0\)

=>(x-18)(x+2)=0

=>x=18 hoặc x=-2

ĐKXĐ:\(x\ne\pm6\)

\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)

\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)

\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)

\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)

\(\Leftrightarrow-4,5x^2+72x+162=0\)

\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)

\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(S=\left\{100\right\}\)

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

lỗi

x + 4/9 + 8/18 + 12/27 + 16/36 = 16/9

 x + (4/9+12/27)+(8/12+16/36) = 16/9

 x +       8/9         +     10/9        = 16/9

  x +               2                         = 16/9

  x                                              = 16/9 - 2

  x                                               =  -2/18  

x+4/9+4/9+4/9+4/9=16/9
x+4/9.4=16/9
x+16/9=16/9
-->    x = 0

a.

\(\dfrac{5x-17}{14}+\dfrac{x-3}{26}>\dfrac{29-9x}{91}\)

\(\Leftrightarrow13\left(5x-17\right)+7\left(x-3\right)>2\left(29-9x\right)\)

\(\Leftrightarrow65x-221+7x-21>58-18x\)

\(\Leftrightarrow65x+7x+18x>58+21+221\)

\(\Leftrightarrow90x>300\)

\(\Leftrightarrow x>\dfrac{10}{3}\)

b)

\(\dfrac{8x-1}{9}+\dfrac{3x-2}{4}< \dfrac{43+8x}{12}+\dfrac{35x}{36}\)

\(\Leftrightarrow4\left(8x-1\right)+9\left(3x-2\right)< 3\left(43+8x\right)+35x\)

\(\Leftrightarrow32x-4+27x-18< 129+24x+35x\)

\(\Leftrightarrow32x+27x-24x-35x< 129+18+4\)

\(\Leftrightarrow0x< 151\) ( luôn đúng)

Vậy bất pt vô số nghiệm

c) \(\dfrac{x-2}{5}+\dfrac{2\left(x+1\right)}{3}>\dfrac{13x-8}{15}\)

\(\Leftrightarrow\dfrac{3x-6}{15}+\dfrac{10x+10}{15}>\dfrac{13x-8}{15}\)

\(\Rightarrow\)3x+10x-6+10>13x-8

\(\Leftrightarrow\)13x-13x+8+4>0

\(\Leftrightarrow0x+12>0\) (vô nghiệm)