`8/(3x-1)=3x+1`                 ĐKXĐ : `x \ne 1/3`

`<=>8=(3x+1)(3x-1)`

`<=>8=9x^2-1`

`<=>9x^2-9=0`

`<=>x^2-1=0`

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

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

`<=>` \( \left[\begin{array}{} x=1 \ \rm(tm)\\ x=-1 \ \rm(tm) \end{array} \right.\)

Vậy `S={\pm 1}`

Em cảm ơn cô ạ

a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`

`=x^3-3^3-(9x^3+27x^2-9x-27)`

`=x^3-3^3-9x^3-27x^2+9x+27`

`=-8x^3-27x^2+9x`

b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`

`=x^3-2^3-x(x^2-9)`

`=x^3-8-x^3+9x`

`=9x-8`

`(x-3)^2+x(1-x)=3`

`<=>x^2-6x+9+x-x^2-3=0`

`<=>-5x+6=0`

`<=>5x=6`

`<=>x=6/5`

Vậy `S={6/5}`

`8/(3x-1)=3x+1`                  ĐKXĐ : `x \ne 1/3`

`<=>8=(3x+1)(3x-1)`

`<=>3x^2-1=8`

`<=>3x^2=9`

`<=>x^2=3`

`<=>x=\pm \sqrt{3}` ( tm ĐKXĐ )

Vậy `S={\pm \sqrt{3}}`

`\sqrt{2x-5}-\sqrt{8x-20}=5`           ĐKXĐ : `x>=5/2`

`<=>\sqrt{2x-5}-\sqrt{4(2x-5)}=5`

`<=>\sqrt{2x-5}-2\sqrt{2x-5}=5`

`<=>-\sqrt{2x-5}=5`

`<=>\sqrt{2x-5}=-5` ( vô lí )

Vậy pt đã cho vô nghiệm

`2.|x-3,5|=1/2`

`->|x-3,5|=0,5:2`

`->|x-3,5|=0,25`

`->` \(\left[\begin{array}{} x-3,5=0,25\\ x-3,5=-0,25 \end{array} \right.\)

`->` \(\left[\begin{array}{} x=3,75\\ x=3,25\end{array} \right.\)

`Δ=1^2-4.1.(-7)=29>0`

`->\sqrt{Δ}=\sqrt{29}`

Vậy phương trình có 2 nghiệm phân biệt

`x_1=(-1+\sqrt{29})/2`

`x_2=(-1-\sqrt{29})/2`