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: \(S=\left\{-\dfrac{8}{15}\right\}\)

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.

giúp mình với ạ câu nào cũng được

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

a: =>(x-2)(2x+5)=0

=>x-2=0 hoặc 2x+5=0

=>x=2 hoặc x=-5/2

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

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

=>x^2+3x=x^2-1

=>3x=-1

=>x=-1/3

\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)

\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\)              ( ĐKXĐ: \(x\ne-1;x\ne1\) )

\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\)  ( nhận )

Vậy phương trình có tập nghiệm S = \(\left\{-\dfrac{1}{3}\right\}\)

\(\dfrac{3x}{2x-4}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\left(đk:x\ne0;x\ne2\right)\)

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

\(\Leftrightarrow3x^2-2x+4=4\)

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

\(\Leftrightarrow x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)

Vậy nghiệm của PT là \(S=\left\{\dfrac{2}{3}\right\}\)

Với \(x\ne0,x\ne2\)

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

<=>\(\dfrac{3x^2-\left(2\left(x-2\right)\right)}{2x\left(x-2\right)}=\dfrac{4}{2x\left(x-2\right)}\)

<=>\(3x^2-2x+4=4\)

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

<=>\(x\left(3x-2\right)=0\)

<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

`(3x)/(2x-4) - 1/x = 2/(x^2 - 2x)` (1)

ĐKXĐ: `x ne 2; x ne 0` 

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

`<=> (3x^2)/[2x(x-2)] - [1.2(x-2)]/[2x(x-2)] = (2.2)/[2x(x-2)]`

`<=> (3x^2 - 2x +4)/[2x(x-2)] = 4/[2x(x-2)]`

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

`<=> 3x^2 - 2x = 0`

`<=> x(3x -2) =0`

`<=>`\(\left[ \begin{array}{l}x=0\\3x-2=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x= \dfrac{2}{3}\end{array} \right.\)

Loại trường hợp `x= 0` vì không thỏa mãn ĐKXĐ

Vậy tập nghiệm của phương trình là `S= { 2/3}`

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)