Question

Answer 1

\(\sqrt{\left(5x+1\right)^2}=\dfrac{6}{7}\)
\(\Leftrightarrow5x+1=\dfrac{6}{7}\)
\(\Rightarrow5x=-\dfrac{1}{7}\)
\(\Rightarrow x=-\dfrac{1}{7}.\dfrac{1}{5}=-\dfrac{1}{35}\)

Answer 2

`sqrt{(5x+1)^2}=6/7`

`|5x+1|=6/7`

`=>[(5x+1=6/7),(5x+1=-6/7):}`

`=>[(5x=6/7-1=-1/7),(5x=-6/7-1=-13/7):}`

`=>[(x=-1/7 :5 =-1/35),(x=-13/7 :5 =-13/35):}`

Vậy `x in {-1/35;-13/35}`