Ta có: \(\frac{1}{\left(3x+1\right)^2}=\frac{4}{9}\Rightarrow\left(3x+1\right)^2=\frac{9}{4}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{3}{2}\\3x+1=-\frac{3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}\\3x=-\frac{5}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}}\)
Vậy x = 1/6 , x = -5/6
Ta nhân chéo nha :
\(\frac{1}{\left(3x+1\right)^2}=\frac{4}{9}\)
\(=>\left(3x+1\right)^2=\frac{4}{9}\)
\(=>\sqrt{\left(3x+1\right)^2}=\sqrt{\left(\frac{4}{9}\right)}\)
\(=>3x+1=\frac{2}{3}\)
\(=>3x=-\frac{1}{3}\)
\(=>x=\frac{-1}{9}\)
Hay lắm
