Đáp án:
c. x=9
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge \dfrac{3}{2}\\
\sqrt {4{x^2} - 9} = 2\sqrt {2x + 3} \\
\to 4{x^2} - 9 = 4\left( {2x + 3} \right)\\
\to 4{x^2} - 8x - 17 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{2 + \sqrt {21} }}{2}\left( {TM} \right)\\
x = \dfrac{{2 - \sqrt {21} }}{2}\left( l \right)
\end{array} \right.\\
b.DK:x \ge 13\\
2.\sqrt {\dfrac{{x - 1}}{4} - 3} = 2\sqrt {\dfrac{{4x - 4}}{9} - \dfrac{1}{3}} \\
\to 2.\sqrt {\dfrac{{x - 1 - 12}}{4}} = 2\sqrt {\dfrac{{4x - 4 - 3}}{9}} \\
\to 2.\dfrac{{\sqrt {x - 13} }}{2} = 2.\dfrac{{2\sqrt {x - 7} }}{3}\\
\to \sqrt {x - 13} = \dfrac{4}{3}\sqrt {x - 7} \\
\to x - 13 = \dfrac{{16}}{9}\left( {x - 7} \right)\\
\to \dfrac{7}{9}x = - \dfrac{5}{9}\\
\to x = - \dfrac{5}{7}\left( l \right)\\
\to x \in \emptyset \\
c.DK:x \ge 5\\
\sqrt {4x - 20} + \sqrt {x - 5} - \dfrac{1}{3}\sqrt {9x - 45} = 4\\
\to 2\sqrt {x - 5} + \sqrt {x - 5} - \dfrac{1}{3}.3\sqrt {x - 5} = 4\\
\to \left( {2 + 1 - 1} \right)\sqrt {x - 5} = 4\\
\to 2\sqrt {x - 5} = 4\\
\to \sqrt {x - 5} = 2\\
\to x - 5 = 4\\
\to x = 9
\end{array}\)
