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: Sửa đề: 2/x+2

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

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`

`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`

`<=>-x-1-x+3=x^2+x-x^2+2x-1`

`<=>-2x+2=3x-1`

`<=>5x=3`

`<=>x=3/5`

Vậy `S={3/5}`

`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`

`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`

`<=>x+3-6x+12+6=0`

`<=>-5x+21=0`

`<=>x=21/5`

Vậy `S={21/5}`

a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

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

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

Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)

\(\Leftrightarrow3x-1=-2x+2\)

\(\Leftrightarrow3x+2x=2+1\)

\(\Leftrightarrow5x=3\)

hay \(x=\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)

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)

Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8+20x-12=0\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

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

=>4x-12x-6+x=0

=>-7x=6

hay x=-6/7

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

\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)

\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)

\(\Leftrightarrow4x-12x-6=-x\)

\(\Leftrightarrow4x-12x-6+x=0\)

\(\Leftrightarrow-7x-6=0\)

\(\Leftrightarrow x=-\dfrac{6}{7}\)

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

=>4x-12x-6+x=0

=>-7x=6

hay x=-6/7

a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)

ĐKXĐ \(x-1\ne0\) hoặc \(x+3\ne0\)

\(\Rightarrow x\ne1\) và \(x\ne-3\)

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

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)

\(\Leftrightarrow9x-x+2x-5x-3x+x=3-5-3-4\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\) (không thỏa ĐK)

Vậy PTVN

b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ: \(x-3\ne0\Rightarrow x\ne3\)

\(x+3\ne0\Rightarrow x\ne-3\)

\(2x+7\ne0\Rightarrow2x\ne-7\Rightarrow x\ne\dfrac{-7}{2}\)

\(\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2+3x-3x-9=12x+42\)

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

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

\(\left\{{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\left(KTĐK\right)\\x=-4\left(TĐK\right)\end{matrix}\right.\)

Vậy S={-4}

a) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\) ( đk: x ≠ 1 ; x ≠ -3 )

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+3x-x-3-4\)

\(\Leftrightarrow3x=-9\)

\(\Rightarrow x=-3\left(KTM\right)\)

S = ∅

b) \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{\left(x-3\right)\left(x+3\right)}\)

( đk: x ≠ ± 3 ; x ≠ \(\dfrac{-7}{2}\) )

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\)

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

\(\Leftrightarrow x^2+3x-4x-12=0\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)

S = \(\left\{4\right\}\)