\(\sqrt{\left(2x-1\right)^2}=9\)
\(\Leftrightarrow\left|2x-1\right|=9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=9\\2x-1=-9\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x=10\\2x=-8\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)
\(\sqrt{\left(2x-1\right)^2}=9\)
<=> (2x - 1)2 = 81 (2 vế không âm)
<=> 4x2 - 4x + 1 = 81
<=> 4x2 - 4x - 80 = 0
<=> 4x2 - 16x + 20x - 80 = 0
<=> 4x.(x - 4) + 20.(x - 4) = 0
<=> (x - 4).(4x - 20) = 0
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\4x+20=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\4x=-20\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-5\end{array}\right.\)
Vậy \(x\in\left\{4;-5\right\}\)
