Đáp án:
a. \(x = \dfrac{8}{9}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0\\
3\sqrt {2x} - 2.2.\sqrt {2x} + 4\sqrt {2x} = 4\\
\to \left( {3 - 4 + 4} \right)\sqrt {2x} = 4\\
\to 3\sqrt {2x} = 4\\
\to \sqrt {2x} = \dfrac{4}{3}\\
\to 2x = \dfrac{{16}}{9}\\
\to x = \dfrac{8}{9}\left( {TM} \right)\\
b.DK:x \ge 2\\
2\sqrt {x - 2} + \dfrac{1}{2}.2\sqrt {x - 2} + 4.\dfrac{1}{8}.\sqrt {x - 2} = 2\\
\to \left( {2 + 1 + \dfrac{1}{2}} \right)\sqrt {x - 2} = 2\\
\to \sqrt {x - 2} = \dfrac{4}{7}\\
\to x - 2 = \dfrac{{16}}{{49}}\\
\to x = \dfrac{{114}}{{49}}\left( {TM} \right)\\
c.DK:x \ge 5\\
\sqrt {x - 5} + 5.\dfrac{1}{5}.\sqrt {x - 5} + \dfrac{1}{2}.2.\sqrt {x - 5} = 7\\
\to \left( {1 + 1 + 1} \right)\sqrt {x - 5} = 7\\
\to \sqrt {x - 5} = \dfrac{7}{3}\\
\to x - 5 = \dfrac{{49}}{9}\\
\to x = \dfrac{{94}}{9}\left( {TM} \right)
\end{array}\)
