`a)(x-1)^2=9`
`→(x-1)^2=(±3)^2`
`→` \(\left[ \begin{array}{l}x-1=3\\x-1=-3\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=4\\x=-2\end{array} \right.\)
Vậy `x∈{4;-2}`
`b)(2x+3)^2=25`
`→(2x+3)^2=(±5)^2`
`→` \(\left[ \begin{array}{l}2x+3=5\\2x+3=-5\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=2\\2x=-8\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=1\\x=-4\end{array} \right.\)
Vậy `x∈{1;-4}`
`d)x^2=0,04`
`→x^2=1/25`
`→x^2=(±1/5)^2`
`→` \(\left[ \begin{array}{l}x=\frac{1}{5}\\x=\frac{-1}{5}\end{array} \right.\)
Vậy `x∈{1/5;-1/5}`
`e)(x+1)^2=16`
`→(x+1)^2=(±4)^2`
`→` \(\left[ \begin{array}{l}x+1=4\\x+1=-4\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
Vậy `x∈{3;-5}`
`g)(2x-1)^2=64`
`→(2x-1)^2=(±8)^2`
`→` \(\left[ \begin{array}{l}2x-1=8\\2x-1=-8\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=9\\2x=-7\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=\frac{9}{2}\\x=\frac{-7}{2}\end{array} \right.\)
Vậy `x∈{9/2;-7/2}`
