Tìm x, biết: a) 2 căn bậc hai của 9x − 27 − 1/5 căn bậc hai của 25x − 75 − 1/7 căn bậc hai của 49x − 147 = 20.

a) \(2\sqrt {9x - 27}  - \frac{1}{5}\sqrt {25x - 75}  - \frac{1}{7}\sqrt {49x - 147}  = 20\)

         ĐKXĐ: \(x \ge 3\)

\(\begin{array}{l}2\sqrt {9x - 27}  - \frac{1}{5}\sqrt {25x - 75}  - \frac{1}{7}\sqrt {49x - 147}  = 20\\2\sqrt {9\left( {x - 3} \right)}  - \frac{1}{5}\sqrt {25\left( {x - 3} \right)}  - \frac{1}{7}\sqrt {49\left( {x - 3} \right)}  = 20\\6\sqrt {x - 3}  - \sqrt {x - 3}  - \sqrt {x - 3}  = 20 \Leftrightarrow 4\sqrt {x - 3}  = 20\\\sqrt {x - 3}  = 5 \Leftrightarrow x - 3 = 25\end{array}\)

\(x = 28\) (TMĐK)

Vậy \(x = 28.\)

b)\(\frac{{3\sqrt x  - 5}}{2} - \frac{{2\sqrt x  - 7}}{3} + 1 = \sqrt x \)

ĐKXĐ: \(x \ge 0\)

\(\begin{array}{l}\frac{{3\sqrt x  - 5}}{2} - \frac{{2\sqrt x  - 7}}{3} + 1 = \sqrt x \\\frac{{\left( {3\sqrt x  - 5} \right).3}}{6} - \frac{{\left( {2\sqrt x  - 7} \right).2}}{6} + \frac{6}{6} = \frac{6}{6}\sqrt x \\9\sqrt x  - 15 - 4\sqrt x  + 14 + 6 = 6\sqrt x \\9\sqrt x  - 4\sqrt x  - 6\sqrt x  = 15 - 14 - 6\\ - \sqrt x  =  - 5\end{array}\)

\(\sqrt x  = 5\)

\(x = 25\) (TMĐK)

Vậy \(x = 25.\)

c)\(\sqrt {9{x^2} + 45}  - \frac{1}{{12}}\sqrt {16{x^2} + 80}  + 3\sqrt {\frac{{{x^2} + 5}}{{16}}}  - \frac{1}{4}\sqrt {\frac{{25{x^2} + 125}}{9}}  = 9;\)

 \[\begin{array}{l}\sqrt {9{x^2} + 45}  - \frac{1}{{12}}\sqrt {16{x^2} + 80}  + 3\sqrt {\frac{{{x^2} + 5}}{{16}}}  - \frac{1}{4}\sqrt {\frac{{25{x^2} + 125}}{9}}  = 9\\\sqrt {9\left( {{x^2} + 5} \right)}  - \frac{1}{{12}}\sqrt {16\left( {{x^2} + 5} \right)}  + 3\frac{{\sqrt {{x^2} + 5} }}{{\sqrt {16} }} - \frac{1}{4}\frac{{\sqrt {25\left( {{x^2} + 5} \right)} }}{{\sqrt 9 }} = 9\\3\sqrt {{x^2} + 5}  - \frac{1}{3}\sqrt {{x^2} + 5}  + \frac{3}{4}\sqrt {{x^2} + 5}  - \frac{5}{{12}}\sqrt {{x^2} + 5}  = 9 \Leftrightarrow 3\sqrt {{x^2} + 5}  = 9\\\sqrt {{x^2} + 5}  = 3\\{x^2} + 5 = 9\\{x^2} = 4\end{array}\]

\(x =  - 2\) hoặc \(x = 2.\)

Vậy \(x =  - 2\) hoặc \(x = 2.\)

d) \(\sqrt {4,5x}  + \sqrt {50x}  - \sqrt {32x}  + \sqrt {72x}  - 5\sqrt {\frac{x}{2}}  - 12 = 0.\)

ĐKXĐ: \(x \ge 0\)

\(\begin{array}{l}\sqrt {4,5x}  + \sqrt {50x}  - \sqrt {32x}  + \sqrt {72x}  - 5\sqrt {\frac{x}{2}}  - 12 = 0\\\sqrt {\frac{{9x}}{2}}  + \sqrt {{5^2}.2x}  - \sqrt {{4^2}.2x}  + \sqrt {{6^2}.2x}  - 5\sqrt {\frac{x}{2}}  - 12 = 0\\\frac{3}{2}\sqrt {2x}  + 5\sqrt {2x}  - 4\sqrt {2x}  + 6\sqrt {2x}  - \frac{5}{2}\sqrt {2x}  = 12 \Leftrightarrow 6\sqrt {2x}  = 12\\\sqrt {2x}  = 2\\2x = 4\end{array}\)

\(x = 2\) (TMĐK)

Vậy \(x = 2.\)