`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: \(S=\left\{\dfrac{3}{5}\right\}\)

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

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}

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

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

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

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

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

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

\(\Leftrightarrow x=3\)

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

`20((x-2)/(x+1))^2-5((x+2)/(x-1))^2+48(x^2-4)/(x^2-1)=0(x ne +-1)`

Đặt `(x-2)/(x+1)=a,(x+2)/(x-1)=b`

`pt<=>20a^2-5b^2+48ab=0`

`<=>20a^2+48ab-5b^2=0`

`<=>20a^2-2ab+50ab-5b^2=0`

`<=>2a(a-10b)+5b(10a-b)=0`

`<=>(a-10b)(2a+5b)=0`

Đến đây dễ rồi bạn tự giải tiếp.

ĐKXĐ: x \(\ne\)\(\pm\)1

Ta có: \(20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)^2+48\cdot\dfrac{x^2-4}{x^2-1}=0\)

Đặt: \(\dfrac{x-2}{x+1}=a\) ; \(\dfrac{x+2}{x-1}=b\)

=> ab = \(\dfrac{x^2-4}{x^2-1}\)

Do đó, ta có pt mới: 20a² - 5b² + 48ab = 0

<=> 20a² + 50ab - 2ab - 5b² = 0

<=> (10a - b)(2a + 5b) = 0

<=> \(\left[{}\begin{matrix}10a=b\\2a=-5b\end{matrix}\right.\)

TH1: 10a = b => \(10\cdot\dfrac{x-2}{x+1}=\dfrac{x+2}{x-1}\)

<=> 10(x - 2)(x - 1) = (x + 2)(x + 1)

<=> 10x² - 30x + 20 = x² + 3x + 2

<=> 9x² - 33x + 18 = 0

<=> 9x² - 27x - 6x + 18 = 0

<=> (9x - 6)(x - 3) = 0

<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{2}{3}\end{matrix}\right.\)(tm)

TH2: \(2a=-5b\)=> \(2\cdot\dfrac{x-2}{x+1}=-5\cdot\dfrac{x+2}{x-1}\)

=> (2x - 4)(x - 1) = (-5x - 10)(x + 1)

<=> 2x² - 6x + 4 = -5x² - 15x - 10

<=> 7x² + 9x + 14 = 0

=> pt vn

mình nhầm mẫu nhé :v mình làm lại 

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

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

Đề sai rồi bạn

\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)

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

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

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

\(\Leftrightarrow2\left(x^2-1\right)=2\left(x^2+4x+4\right)\)

\(\Leftrightarrow2x^2+8x+8-2x^2+2=0\)

=>8x+10=0

hay x=-5/4