Question

Tim x biet

\(\left(2x-1\right)^2-2=30\)

Answer 1

ta có : \(\left(2x-1\right)^2-2=30\Leftrightarrow\left(2x-1\right)^2=32\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{32}\\2x-1=-\sqrt{32}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1+\sqrt{32}\\2x=1-\sqrt{32}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{32}}{2}\\x=\dfrac{1-\sqrt{32}}{2}\end{matrix}\right.\) vậy .............................

Answer 2

\(\left(2x-1\right)^2-2=30\)

\(\Leftrightarrow\left(2x-1\right)^2=30+2\)

\(\Leftrightarrow\left(2x-1\right)^2=32\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=\sqrt{32}\\2x-1=-\sqrt{32}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{32}+1\\2x=-\sqrt{32+1}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{32}+1}{2}\\x=\dfrac{-\sqrt{32}+1}{2}\end{matrix}\right.\)

Vậy ......