a, ta có (5x+1)²=`36/49`
⇔\(\left[ \begin{array}{l}(5x+1)²=(6/7)²\\(5x+1)²=(-6/7)²\end{array} \right.\)
⇔\(\left[ \begin{array}{l}5x+1=6/7\\5x+1=-6/7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}5x=-1/7\\5x=-13/7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1/35\\x=-13/35\end{array} \right.\)