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}

\(DKXD:x\ne\pm1\\ pt\Rightarrow2x^2+x+1+2x-2=x^2-1\\ \Leftrightarrow x^2+3x=0\\ \Leftrightarrow x\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\left(N\right)\\ \Rightarrow S=\left\{0;-3\right\}\)

ĐKXĐ : \(x\ne\pm1\)

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

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

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

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

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy ...

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

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

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

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

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

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

Suy ra: \(-\left(x^2-2x+1\right)-x-1+x^2-x+2x-2=0\)

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

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(nhận)

Vậy: S={2}

ĐKXĐ: \(x\ne\pm1\)

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

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

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

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\)(loại)

Vậy phương trình vô nghiệm

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

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

\(\Leftrightarrow4x-2-3x+3+x+1-6=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

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

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

=>7x=-1

hay x=-1/7

`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 :....

ĐKXĐ: x<>0; y<>0

\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{6}{x}+\dfrac{3}{y}=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{x}=4\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\\dfrac{1}{y}=-1-\dfrac{2}{x}=-1-2:\dfrac{-1}{4}=-1+8=7\end{matrix}\right.\)

=>x=-1/4 và y=1/7

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\end{matrix}\right.\) 

Hệ phương trình trở thành \(\left\{{}\begin{matrix}5a+3b=1\\2a+b=-1\end{matrix}\right.\)

 \(\Rightarrow\left\{{}\begin{matrix}b=-1-2a\\5a+3\left(-1-2a\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1-2a\\-a-3=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=-4\\b=-1-2.\left(-4\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}a=\dfrac{1}{x}=-4\\b=\dfrac{1}{y}=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\y=\dfrac{1}{7}\left(tm\right)\end{matrix}\right.\)

Vậy HPT có nghiệm \(x=-\dfrac{1}{4}\) và \(y=\dfrac{1}{7}\)

Ta có: 

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

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)