a, <=> \(\left(\sqrt{2x-1}\right)^2=3^2\)
<=> 2x - 1 = 9
<=> 2x = 8
<=> x = 4
Vậy ...
b, <=> | 3x + 2 | = 5
<=> 3x + 2 = 5 hoặc 3x + 2 = -5
TH1 : 3x + 2 = 5
<=> 3x = 3
<=> x = 3
TH2 : 3x + 2 = -5
<=> 3x = -7
<=> x= \(-\dfrac{7}{3}\)
Vậy .....
c, <=> \(\sqrt{\left(x-2\right)^2}=3\)
<=> | x - 2 | = 3
<=> x - 2 = 3 hoặc x - 2 = -3
TH1: x - 2 = 3
<=> x = 5
TH2: x - 2 = -3
<=> x = -1
Vậy ...
c: \(\Leftrightarrow\left|x-2\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d, <=> \(\sqrt{x-2}-2\sqrt{x-2}+6\sqrt{x-2}=10\)
<=> 5\(\sqrt{x-2}\) = 10
<=> \(\sqrt{x-2}\) = 2
<=> \(\left(\sqrt{x-2}\right)^2\) = \(2^2\)
<=> x - 2 = 4
<=> x = 6
Vậy...