Câu hỏi/bài tập:
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) \(2 - \sqrt[3]{{5 - x}} = 0\)
Bình phương (lập phương) 2 vế.
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) \(\sqrt {\frac{x}{5}} = 4\)
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\(\begin{array}{l}\sqrt {\frac{x}{5}} = 4\\\frac{x}{5} = 16\\x = 80(tm)\end{array}\)
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 \(x = - 3.\)