Câu hỏi
Tìm \(x\) biết:
\(a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3\)
- A \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{- 7 \right\}.
\end{array}\) - B \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 7 \right\}.
\end{array}\) - C \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 2 \right\}.
\end{array}\) - D \(\begin{array}{l}
a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\
b)\,\,S = \left\{ 2;\,7 \right\}.
\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2 \Leftrightarrow \left\{ \begin{array}{l}2x - 1 \ge 0\\x - 1 > 0\\\sqrt {2x - 1} = 2\sqrt {x - 1} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{1}{2}\\x > 1\\2x - 1 = 4\left( {x - 2} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x > 1\\2x - 1 = 4x - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 1\\2x = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 1\\x = \frac{3}{2}\end{array} \right. \Leftrightarrow x = \frac{3}{2}.\end{array}\)
Vậy phương trình có nghiệm \(x = \frac{3}{2}.\)
\(\begin{array}{l}b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3 \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 4 \ge 0\\x - 2 > 0\\\sqrt {{x^2} - 4} = 3\sqrt {x - 2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge 2\\x \le - 2\end{array} \right.\\x > 2\\{x^2} - 4 = 9\left( {x - 2} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x > 2\\{x^2} - 9x + 14 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 2\\\left( {x - 2} \right)\left( {x - 7} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 2\\\left[ \begin{array}{l}x = 2\\x = 7\end{array} \right.\end{array} \right. \Leftrightarrow x = 7.\end{array}\)
Vậy phương trình có nghiệm duy nhất \(x = 7.\)
Câu hỏi trước
