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

Suy ra: \(3x-6-x+1=2x+4\)

\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)

2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)

Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)

\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)

\(\Leftrightarrow5x=16\)

hay \(x=\dfrac{16}{5}\)

ĐKXĐ: `x ne 1`

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

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

`<=> (5x-2)/[2(1-x)] + (2x - 2x^2 +1 - x)/[2(1-x)] = (2-2x)/[2(1-x)] - (2x^2 +2x -6)/[2(1-x)]`

`=> 5x-2 + x - 2x^2 +1  = 2-2x - 2x^2 - 2x + 6`

`<=> (5x + x + 2x+2x) - (2x^2 + 2x^2) = 2 + 6 - 1 +2`

`<=> 10x = 9`

`<=> x= 9/10` (thỏa mãn ĐKXĐ)

Vậy tập nghiệm của phương trình là `S = { 9/10}`

Dài qué gửi riêng qua mes thôi :^^

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

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

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

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

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

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

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

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

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

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

Đặt \(x^2-2x+2=t\)

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

\(\Rightarrow x^2-2x+4=t+2\)

\(pt\Leftrightarrow \frac{1}{t}+\frac{2}{t+1}=\frac{6}{t+2}\)

\(\Rightarrow (t+1)(t+2)+2t(t+2)=6t(t+1)\)

\(\Leftrightarrow t^2+3t+2+2t^2+4t=6t^2+6t\)

\(\Leftrightarrow 3t^2-t-2=0\)

TH1\( : t=1\)

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

\(\Leftrightarrow x=1\)

TH2:\(t=\frac{-2}{3}\) (loại)

Vậy \(x=1\)

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

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

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)

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

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

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

đkxđ \(x\ne0;2\)

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

mà \(x\ne0\Rightarrow x+1=0\\ \Leftrightarrow x=-1\)

=2/x(x-2) + x-2/x(x-2)=x(x+2)/x(x-2)

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

<=>4 -x -x^2=0

<=>-x(1+x)=4

b) Đặt \(x^2+2x+3=a\)(a>0)

Ta có: \(\dfrac{x^2+2x+7}{\left(x+1\right)^2+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+1+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+3}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{a+4}{a}=a+1\)

\(\Leftrightarrow a^2+a=a+4\)

\(\Leftrightarrow a^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-2\left(loại\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

ĐKXĐ của cả 2 pt trên đều là `x in RR`

`a,1/(x^2-2x+2)+2/(x^2-2x+3)=6/(x^2-2x+4)`

Đặt `a=x^+2x+3(a>=2)` ta có:

`1/(a-1)+2/a=6/(a+1)`

`<=>a(a+1)+2(a-1)(a+1)=6a(a-1)`

`<=>a^2+a+2(a^2-1)=6a^2-6a`

`<=>a^2+a+2a^2-2=6a^2-6a`

`<=>3a^2-5a+2=0`

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

`<=>3a(a-1)-2(a-1)=0`

`<=>(a-1)(3a-2)=0`

`a>=2=>a-1>=1>0`

`a>=2=>3a-2>=4>0`

Vậy pt vô nghiệm

`(x^2+2x+7)/((x+1)^2+2)=x^2+2x+4`

`<=>(x^2+2x+7)=(x^2+2x+4)(x^2+2x+3)`

Đặt `a=x^2+2x+3(a>=2)`

`pt<=>a+4=a(a+1)`

`<=>a^2+a=a+4`

`<=>a^2=4`

`<=>a=2` do `a>=2`

`<=>(x+1)^2+2=2`

`<=>(x+1)^2=0`

`<=>x=-1`

Vậy `S={-1}`