\(\left(2x-3\right)^2=26\Leftrightarrow\left\{{}\begin{matrix}2x-3=\sqrt{26}\\2x-3=-\sqrt{26}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=\sqrt{26}+3\\2x=-\sqrt{26}+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt{26}+3}{2}\\x=\dfrac{3-\sqrt{26}}{2}\end{matrix}\right.\) vậy \(x=\dfrac{\sqrt{26}+3}{2};x=\dfrac{3-\sqrt{26}}{2}\)
\(\)\(\left(2x-3\right)^2=26\)
\(\Rightarrow\left(2x-3\right)^2=\pm\sqrt{26}^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\sqrt{26}\\2x-3=-\sqrt{26}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{26}+3\Rightarrow x=\dfrac{\sqrt{26}+3}{2}\\2x=-\sqrt{26}+3\Rightarrow x=\dfrac{-\sqrt{26}+3}{2}\end{matrix}\right.\)
Đề sai phải ko?
\(\left(2x-3\right)^2=25\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=5^2\\\left(2x-3\right)^2=\left(-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\) \(\left(tm\right)\)
Vậy .................
