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 \(\begin{array}{l} a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\ b)\,\,S = \left\{ 2;\,7 \right\}. \end{array}\)

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}\)

Hướng dẫn Chọn đáp án là: B

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 \(x = 7.\)