Tìm `x:` $\sqrt{(3x +1)^2}$ `= 6/7`

Đáp án + Giải thích các bước giải:

$\sqrt{(3x+1)^2}$ `=6/7`

`=>(\sqrt{(3x+1)^2})^2=(6/7)^2`

`=>(3x+1)^2=36/49`

`=>(3x+1)^2=(±6/7)^2`

`=>3x+1=6/7` hoặc `3x+1=-6/7`

`TH1:3x+1=6/7`

`=>3x=6/7-1`

`=>3x=-1/7`

`=>x=(-1/7):3`

`=>x=-1/21`

`TH2:3x+1=-6/7`

`=>3x=(-6/7)-1`

`=>3x=-13/7`

`=>x=(-13/7):3`

`=>x=-13/21`

Vậy `x∈{-1/21;-13/21}`