giúp mình với ạ câu nào cũng được

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-3\\x\ne1\end{matrix}\right.\)

\(\dfrac{2x}{x+3}-\dfrac{x}{x-1}=\dfrac{-9x-3}{x^2+2x-3}\\ \Leftrightarrow\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{-\left(9x+3\right)}{\left(x-1\right)\left(x+3\right)}\\ \Leftrightarrow\dfrac{2x^2-2x}{\left(x-1\right)\left(x+3\right)}-\dfrac{x^2+3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{9x+3}{\left(x-1\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{2x^2-2x-x^2-3x+9x+3}{\left(x-1\right)\left(x+3\right)}=0\\ \Rightarrow x^2+4x+3=0\\ \Leftrightarrow\left(x+1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

đk : x khác -3 ; 1 

\(2x^2-2x-x^2-3x=-9x-3\Leftrightarrow x^2+4x+3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=-3\left(ktm\right);x=-1\left(tm\right)\)

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

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

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

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

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

a>16-x/4=2x+1/3

<=>3[16-x)=4(2x+1)

<=>48-3x=8x+8

<=>-3x-8x=8-48

<=>-5x=-40

<=>x=8

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)

\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

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

a.\(\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

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

c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{x^2-3x+9}{x-3}\cdot\dfrac{3}{-\left(x^2-3x+9\right)}\)

\(=\dfrac{-3}{x-3}\)

1. 

ĐK: \(x\ne3;x\ne-2\)

\(\dfrac{5}{x-3}+\dfrac{3}{x+2}\le\dfrac{3+2x}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{5\left(x+2\right)+3\left(x-3\right)}{x^2-x-6}\le\dfrac{3+2x}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{8x+1-3-2x}{x^2-x-6}\le0\)

\(\Leftrightarrow\dfrac{6x-2}{x^2-x-6}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}6x-2\ge0\\x^2-x-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\-2< x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{3}\le x< 3\)

TH2: \(\left\{{}\begin{matrix}6x-2\le0\\x^2-x-6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x< -2\)

Vậy ...

2.

ĐK: \(x\ne\pm2\)

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

\(\Leftrightarrow\dfrac{1}{x^2-4}+\dfrac{2\left(x-2\right)+3\left(x+2\right)}{x^2-4}>0\)

\(\Leftrightarrow\dfrac{5x+3}{x^2-4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5x+3>0\\x^2-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}5x+3< 0\\x^2-4< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{5}< x< 2\\x< -2\end{matrix}\right.\)

Vậy ...

8.

\(\sqrt{x^2+6x+9}-2x+1>0\)

\(\Leftrightarrow\sqrt{x^2+6x+9}>2x-1\)

\(\Leftrightarrow\sqrt{x^2+6x+9}>2x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1< 0\\x^2+6x+9\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\ge0\\x^2+6x+9>\left(2x-1\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\\dfrac{1}{2}\le x< 4\end{matrix}\right.\)

Vậy ...