Đề bài

Tìm x, biết: a) \(\frac{1}{2}\sqrt x  - \frac{3}{2}\sqrt {9x}  + 24\sqrt {\frac{x}{{64}}}  =  - 17\) với \(x \ge 0\) b) \(\sqrt {\frac{x}{5}}  = 4\) với \(x \ge 0\) c) \(\sqrt {25{x^2}}  = 10\) d) \(\sqrt {{{\left( {2x - 1} \right)}^2}}  = 3\) e) \(2 - \sqrt[3]{{5 - x}} = 0\)

Lời giải chi tiết

a) \(\frac{1}{2}\sqrt x - \frac{3}{2}\sqrt {9x} + 24\sqrt {\frac{x}{{64}}} = - 17\) \(\begin{array}{l}\frac{1}{2}\sqrt x - \frac{3}{2}\sqrt {9x} + 24\sqrt {\frac{x}{{64}}} = - 17\\\frac{1}{2}\sqrt x - \frac{9}{2}\sqrt x + 3\sqrt x = - 17\\\sqrt x \left( {\frac{1}{2} - \frac{9}{2} + 3} \right) = - 17\\\sqrt x \left( {\frac{1}{2} - \frac{9}{2} + 3} \right) = - 17\\ - \sqrt x = - 17\\\sqrt x = 17\\x = 289(tm)\end{array}\) Vậy \(x = 289\). b) \(\sqrt {\frac{x}{5}} = 4\) \(\begin{array}{l}\sqrt {\frac{x}{5}} = 4\\\frac{x}{5} = 16\\x = 80(tm)\end{array}\) Vậy \(x = 80\). c) \(\sqrt {25{x^2}} = 10\) \(\begin{array}{l}\sqrt {25{x^2}} = 10\\25{x^2} = 100\\{x^2} = 4\end{array}\) \(x = 2\) hoặc \(x = - 2\) Vậy \(x = 2\);\(x = - 2\) d) \(\sqrt {{{\left( {2x - 1} \right)}^2}} = 3\) \({\left( {2x - 1} \right)^2} = 9\) \(2x - 1 = 3\) hoặc \(2x - 1 = - 3\) \(2x = 4\) hoặc \(2x = - 2\) \(x = 2\) hoặc \(x = - 1\) Vậy \(x = 2\);\(x = - 1\) e) \(2 - \sqrt[3]{{5 - x}} = 0\) \(\begin{array}{l}\sqrt[3]{{5 - x}} = 2\\5 - x = 8\\x = - 3\end{array}\) Vậy \(x = - 3.\)