Đáp án:
\(\begin{array}{l}
a)x = 9\\
b)\left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 0\\
x = \dfrac{2}{3}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 5\\
\sqrt {x - 5} = 2\\
\to x - 5 = 4\\
\to x = 9\\
b)\sqrt {{x^2} - 6x + 9} = 5\\
\to {x^2} - 6x + 9 = 25\\
\to {x^2} - 6x - 16 = 0\\
\to \left( {x - 8} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 8\\
x = - 2
\end{array} \right.\\
c)\sqrt {4{x^2} - 4x + 1} = x - 1\\
\to 4{x^2} - 4x + 1 = {x^2} - 2x + 1\left( {DK:x \ge 1} \right)\\
\to 3{x^2} - 2x = 0\\
\to x\left( {3x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{2}{3}
\end{array} \right.\\
d)\sqrt {{x^2} - 4x + 4} = \sqrt {4{x^2} - 12x + 9} \\
\to {x^2} - 4x + 4 = 4{x^2} - 12x + 9\\
\to 3{x^2} - 8x + 5 = 0\\
\to \left( {3x - 5} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = 1
\end{array} \right.
\end{array}\)
