Đáp án:
\(\begin{array}{l}
a)x = 6\\
c)\left[ \begin{array}{l}
x = 10\\
x = - 2
\end{array} \right.\\
b)x = \dfrac{9}{2}\\
d)\left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 2\\
3\sqrt {x - 2} + 2\sqrt {x - 2} = 10\\
\to 5\sqrt {x - 2} = 10\\
\to \sqrt {x - 2} = 2\\
\to x - 2 = 4\\
\to x = 6\\
c)\sqrt {{x^2} - 8x + 16} = 6\\
\to \sqrt {{{\left( {x - 4} \right)}^2}} = 6\\
\to \left| {x - 4} \right| = 6\\
\to \left[ \begin{array}{l}
x - 4 = 6\\
x - 4 = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\\
x = - 2
\end{array} \right.\\
b)DK:x \ge 1\\
2\sqrt {x - 1} = \sqrt {2x + 5} \\
\to 4\left( {x - 1} \right) = 2x + 5\\
\to 2x = 9\\
\to x = \dfrac{9}{2}\\
d)\sqrt[3]{{{x^2} - 9}} = - 2\\
\to {x^2} - 9 = - 8\\
\to {x^2} = 1\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.
\end{array}\)
