Đáp án:

 $\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}$

$⇔ \dfrac{2.(x+5)}{6.(x-2)}-\dfrac{1.3.(x-2)}{6.(x-2)}=\dfrac{(2x-3).3}{6.(x-2)}$

$⇒ 2x+10-3x+6=6x-9$

$⇔ -x+16-6x+9=0$

$⇔ -7x+25=0$

$⇔ -7x=-25$

$⇔ x= \dfrac{25}{7}$

$\text{Vậy S=}$ $\left\{{\dfrac{{25}}{{7}}}\right\}$

$\dfrac{12}{1-9x²}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$ $\text{(ĐK: x $\neq$ ± 1/3)}$

$⇔ \dfrac{12}{(1-3x).(1+3x)}=\dfrac{(1-3x)²}{(1-3x).(1+3x)}-\dfrac{(1+3x)^2}{(1-3x).(1+3x)}$

$⇒ 12= (1-3x)^2-(1+3x)^2$

$⇔ 12= (1-3x-1-3x).(1-3x+1+3x)$

$⇔ 12= -6x.2$

$⇔ 12+12x=0$

$⇔ 12x=-12$

$⇔ x=-1$

$\text{Vậy S={-1}}$

 

Đáp án:

`a)`

ĐKXĐ : `x \ne 2`.

`(x+5)/(3x-6)-1/2=(2x-3)/(2x-2)`

`⇔(x+5)/[3(x-2)]-1/2=(2x-3)/[2(x-2)]`

`⇔[2(x+5)]/[6(x-2)]-[3(x-2)]/[6(x-2)]=[3(2x-3)]/[6(x-2)]`

`⇒2x+10-3x+6=6x-9`

`⇔-x+16=6x-9`

`⇔-x-6x=-9-16`

`⇔-7x=-25`

`⇔x=25/7`   (Thỏa mãn)

Vậy `S={25/7}`

`b)`

ĐKXĐ : `x \ne +-1/3`

`12/(1-9x^2)=(1-3x)/(1+3x)-(1+3x)/(1-3x)`

`⇔12/[(1-3x)(1+3x)]=[(1-3x)(1-3x)]/[(1+3x)(1-3x)]-[(1+3x)(1+3x)]/[(1-3x)(x+3x)]`

`⇒12=(1-3x)^(2)-(1+3x)^2`

`⇔12=1-6x+9x^(2)-1-6x-9x^2`

`⇔12x=-12`

`⇔x=-1`    (Thỏa mãn)

Vậy `S={-1}`