ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)

Vậy \(S\in\left\{\varnothing\right\}\)

ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)

\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)

\(\to x^2+3x+2-5x+10=12+x^2-4\)

\(\leftrightarrow x^2-2x-x^2=12-12-4\)

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

\(\leftrightarrow x=2(\rm KTM)\)

Vậy pt đã cho vô nghiệm \(S=\varnothing\)

\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\Leftrightarrow x\left(x+8\right)=105\)

\(\Leftrightarrow x^2+8x-105=0\)

\(\Leftrightarrow x^2-7x+15x-105=0\)

\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)

ĐKXĐ: ` x ne 1 ; x ne 4`

`(2x+1)/(x^2-5x+4) + 5/(x-1) = 2/(x-4)`

`<=> (2x+1)/[(x-1)(x-4)] + [5(x-4)]/[(x-1)(x-4)] = [2(x-1)]/[(x-1)(x-4)]`

`=> 2x+1 + 5x -20 = 2x-2`

`<=> 5x = 17`

`<=> x= 17/5`(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của phương trình là `S={ 17/5}`

ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)

Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)

\(\Leftrightarrow\dfrac{2x+4}{x.\left(x+4\right)}+\dfrac{2x+4}{\left(x+1\right).\left(x+3\right)}+\dfrac{1}{x+2}=0\)

\(\Leftrightarrow\dfrac{2x+4}{\left(x+2\right)^2-4}+\dfrac{2x+4}{\left(x+2\right)^2-1}+\dfrac{1}{x+2}=0\) (*)

Đặt x + 2 = a \(\left(a\ne0\right)\) 

(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)

\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)

Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)

Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)

\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)

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

\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)

\(\Leftrightarrow4b^4-15b^2+5=0\) (***)

Đặt b² = t > 0

Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)

\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\) 

mà x + 2 = a ; ab = 1 

nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)

Thử lại ta có phương trình có 4 nghiệm như trên

ĐKXĐ: \(x\ne\pm2\)

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

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

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

\(\Leftrightarrow4x=-2\)

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

Vậy...

\(\dfrac{4}{x+2}-\dfrac{1}{x}=\dfrac{x^2-2}{x^2+2x}\left(x\ne0;x\ne-2\right)\)

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

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

\(\Rightarrow4x-\left(x+2\right)=x^2-2\)

\(\Leftrightarrow4x-x-x^2=2-2\)

\(\Leftrightarrow3x-x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=3\left(tm\right)\end{matrix}\right.\)

`1+(x-2)/(1-x)+(2x^2-5)/(x^3-1)=4/(x^2+x+1)(x ne 1)`

`<=>(x^3-1)/(x^3-1)-((x-2)(x^2+x+1))/(x^3-1)+(2x^2-5)/(x^3-1)=(4(x-1))/(x^3-1)`

`<=>x^3-1-(x-2)(x^2+x+1)+2x^2-5=4(x-1)`

`<=>x^3-1-(x^3-x^2-x-2)+2x^2-5=4x-4`

`<=>x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0`

`<=>3x^2-3x+2=0`

`<=>x^2-2/3 x+2/3=0`

`<=>x^2-2.x. 1/3+1/9+5/9=0`

`<=>(x-1/3)^2=-5/9` vô lý

Vậy phương trình vô nghiệm.

ĐKXĐ: \(x\ne1\)

Ta có: \(1+\dfrac{x-2}{1-x}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}\)

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

Suy ra: \(x^3-1-\left(x^3+x^2+x-2x^2-2x-2\right)+2x^2-5=4x-4\)

\(\Leftrightarrow x^3-1-x^3+x^2+x+2+2x^2-5-4x+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

mà 3>0

nên x(x-1)=0

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

Vậy: S={0}

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

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

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

\(\rightarrow\dfrac{13}{12}x=\dfrac{13}{12}\)

\(\rightarrow x=1\)

ĐKXĐ : \(x\ne\pm2\)

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

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

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

\(\Leftrightarrow x^2+3x+1=0\)

\(\Leftrightarrow x=\dfrac{-3\pm\sqrt{5}}{2}\left(TM\right)\)

Vậy ...