\(\left(2x-1\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=8\\2x-1=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy..............
Ta có: (2x - 1)2 = 64
\(\Rightarrow\left[{}\begin{matrix}2x-1=8\\2x-1=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{9}{2};x_2=-\dfrac{7}{2}\)
Ta có:
\(\left(2x-1\right)^2=64\)
Mà
\(64=8^2=\left(-8\right)^2\)
\(\Leftrightarrow2x-1=8\Rightarrow2x=9\Rightarrow x=\dfrac{9}{2}\)
\(\Leftrightarrow2x-1=-8\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
Vậy......
( 2x - 1 )\(^2\) = 64
=> ( 2x -1 )\(^2\) = 8\(^2\) ( Hoặc -8\(^2\) )
Trường hợp 1 :
Nếu ( 2x - 1 )\(^2\) = 8\(^2\)
=> 2x - 1 =8
=> 2x = 9
=> x = 9 : 2
=> x = 4,5
Trường hợp 2 :
Nếu ( 2x -1 ) \(^2\) = ( -8 )\(^2\)
=> 2x ( -1 ) = -8
=> 2x = ( -8 ) + 1
=> 2x = 7
=> x = 7 : 2
=> x = 3,5
Vậy : x = 3,5 hoặc x = 4,5
\(\left(2x-1\right)^2=64\)
\(\Rightarrow\left(2x-1\right)^2=8^2\)
\(\Rightarrow2x-1=\pm8\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=8\\2x-1=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)
\(\left(2x-1\right)^2=64\)
\(\Rightarrow\left(2x-1\right)^2=+8^2=-8^2\)
\(1)2x-1=8\rightarrow2x=9\rightarrow x=4,5\)
2)\(2x-1=-8\rightarrow2x=-7\rightarrow x=(-7)\div2=-3,5\)
\((\) \(2x-1\) \()\) \(^2\) \(=\) \(64\)
=> \((\) \(2x-1\) \()\) \(^2\) \(=\) \(+\) \(8^2\) \(=\) \(-8^2\)
\(2x-1\) \(=\) \(8\) => \(2x=9\) => x \(=\) \(\dfrac{9}{2}\) \(=\) \(4,5\)
\(2x-1=-8\) => \(2x=-7\) => x \(=\) \(-\dfrac{7}{2}\) \(=\) \(-3,5\)
