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

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

\(\Rightarrow x+1-2x=3\)

\(\Leftrightarrow1-x=3\)

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

Vậy phương trình có nghiệm duy nhất \(x=-2\)

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

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

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

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

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

\(\Leftrightarrow x=0\) hoặc x + 1= 0

⇔ x = 0 hoặc x= -1

Vậy phương trình có tập nghiệm là S={0;-1}

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

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

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

Suy ra: \(x+1-2x=3\)

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

\(\Leftrightarrow-x=2\)

hay x=-2(thỏa ĐK)

Vậy: S={-2}

=>(x^2+1)^2+x^2/x*(x^2+1)=5/2

=>\(\dfrac{\left(x^2+1\right)^2+x^2}{x\left(x^2+1\right)}=\dfrac{5}{2}\)

=>\(2\left(x^4+2x^2+1+x^2\right)=5\left(x^3+x\right)\)

=>2x^4+6x^2+2-5x^3-5x=0

=>2x^4-5x^3+6x^2-5x+2=0

=>2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0

=>(x-1)(2x^3-3x^2+3x-2)=0

=>(x-1)(2x^3-2x^2-x^2+x+2x-2)=0

=>(x-1)^2*(2x^2-x+2)=0

=>x-1=0

=>x=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

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

\(\Leftrightarrow-4x=12\)

\(\Rightarrow x=3\)

Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

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

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

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

=>\(2x^2-x+3x+2=x^2+3x+2\)

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

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

=>x(x-1)=0

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

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

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

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

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

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

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

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

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

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

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

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

a: =>10x=3(5-3x)

=>10x=15-9x

=>19x=15

=>x=15/19

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

=>2x^2+2x=x^2-4x+x^2-1=2x^2-4x-1

=>2x=-4x-1

=>6x=-1

=>x=-1/6

c:=>x(x+2)-x+2=2

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

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

d: =>x+1+3x=2

=>4x=1

=>x=1/4

e: =>x(x+1)+x(x-3)=2x

=>x^2+x+x^2-3x=2x

=>2x^2-4x=0

=>x=0(nhận) hoặc x=2(nhận)

f: =>2x+6-4x+12=5

=>-2x=-13

=>x=13/2

\(ĐK:x\ne-1;x\ne1\\ PT\Leftrightarrow\dfrac{\dfrac{2x^2+4x+2-x^2+2x-1}{2\left(x+1\right)\left(x-1\right)}}{\dfrac{x-1+x+1}{x-1}}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2x}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{4x\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow x^2+6x+1=2x\left(x-1\right)\\ \Leftrightarrow x^2+6x+1=2x^2-2x\\ \Leftrightarrow x^2-8x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{17}\left(tm\right)\\x=4-\sqrt{17}\left(tm\right)\end{matrix}\right.\)

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)