\(\left(3x-1\right)^2=44\)
\(x=\dfrac{6\pm\sqrt{\left(-6\right)^2-4.9\left(-43\right)}}{2.9}\)
\(x=\dfrac{1\pm2\sqrt{11}}{3}\)
oi luoi qua :")
\(\left(3x-1\right)^2=144\\\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^2=12^2\\\left(3x-1\right)^2=\left(-12\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-1=12\\3x-1=-12\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=13\\3x=-11\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(\left(3x-1\right)^2=44\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\sqrt{11}\\3x-1=-2\sqrt{11}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{11}+1}{3}\\x=\dfrac{-2\sqrt{11}+1}{3}\end{matrix}\right.\)
Ta có : \(\left(3x-1\right)^2=44\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\sqrt{44}\\3x-1=-\sqrt{44}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\sqrt{44}+1\\3x=1-\sqrt{44}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{44}}{3}\\x=\dfrac{1-\sqrt{44}}{3}\end{matrix}\right.\)
Vậy : \(x=\dfrac{1\pm\sqrt{44}}{3}\)
