$\frac{2x+1}{x-1}$ = $\frac{5(x-1)^2}{x+1}$      Đkxđ : x $\neq$ ± 1

=> (2x+1)(x+1)=$5(x-1)^{2}$

<=> 2$x^{2}$ +2x+x+1=5($x^{2}$ -2x+1)

<=> $2x^{2}$ -$5x^{2}$ +3x+10x+1-5=0

<=> $-3x^{2}$ +13x-4=0

<=> $-3x^{2}$ +12+x-4=0

<=> -3x(x-4)+(x+4 ) =0

<=> (x-4)(-3x+1)=0

<=> \(\left[ \begin{array}{l}x-4=0\\-3x+1=0\end{array} \right.\)

<=> \(\left[ \begin{array}{l}x=4\\x=\frac{1}{3}\end{array} \right.\)

Vậy S = {4; $\frac{1}{3}$ }

 

Đáp án:

 

Giải thích các bước giải:

`(2x+1)/(x-1)=[5(x-1)]/(x+1)`

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

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

`<=>2x^2+3x+1=5x^2-10x+5`

`<=>2x^2-5x^2+3x+10x+1-5=0`

`<=>-3x^2+13x-4=0`

`<=>-3x^2+12x+x-4=0`

`<=>-3x(x-4)+(x-4)=0`

`<=>(x-4)(-3x+1)=0`

`=>x-4=0`

hoặc `-3x+1=0`

`=>x=4`

hoặc `x=1/3`