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}

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

`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: \(S=\left\{\dfrac{3}{5}\right\}\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a ) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\) (1)

ĐKXĐ : x\(\ne1;-2.\)

\(\left(1\right)\Leftrightarrow x+2-7x+7=3\)

\(\Leftrightarrow-6x=-6\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy pt vô nghiệm .

b ) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)

Đặt \(x^2+2x+1=t\) ta được :

\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow6t^2+12t+6t^2+12t+6=7\left(t^2+3t+2\right)\)

\(\Leftrightarrow5t^2+3t-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{8}{5}\end{matrix}\right.\)

Khi t = 1

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

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

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

Khi \(t=-\dfrac{8}{5}\)

\(\Leftrightarrow\left(x+1\right)^2=-\dfrac{8}{5}\) ( vô lí )

Vậy ............

a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10