Tìm x :
a) \(\sqrt {2x - 1} = 5\);
b) \(\sqrt {2x - 1} = \left| { - 3} \right|\);
c) \(\sqrt {{{\left( {2x - 5} \right)}^2}} = 4\);
d) \(\sqrt {{{\left( {3x - 2} \right)}^2}} = \left| { - 2} \right|\);
e) \(\sqrt {{{\left( {x - 2} \right)}^2}} = 2x - 1\).
+) Tìm tập xác định của phương trình.
+) Giải phương trình bằng phương pháp bình phương hai vế.
\(a)\;\;\sqrt {2x - 1} = 5\)
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Điều kiện: \(2x - 1 \ge 0 \Leftrightarrow x \ge \dfrac{1}{2}.\)
\(PT \Leftrightarrow 2x - 1 = 25 \Leftrightarrow 2x = 26 \Leftrightarrow x = 13\;\;\;\left( {tm} \right).\)
Vậy \(x = 13.\)
\(b)\;\sqrt {2x - 1} = \left| { - 3} \right|\)
Điều kiện: \(2x - 1 \ge 0 \Leftrightarrow x \ge \dfrac{1}{2}.\)
\(PT \Leftrightarrow \sqrt {2x - 1} = 3 \Leftrightarrow 2x - 1 = 9 \Leftrightarrow 2x = 10 \Leftrightarrow x = 5\;\;\left( {tm} \right).\)
Vậy \(x = 5.\)
\(\begin{array}{l}c)\;\;\sqrt {{{\left( {2x - 5} \right)}^2}} = 4\\ \Leftrightarrow \left| {2x - 5} \right| = 4\\ \Leftrightarrow \left[ \begin{array}{l}2x - 5 = 4\\2x - 5 = - 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = 9\\2x = 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{9}{2}\\x = \dfrac{1}{2}\end{array} \right..\end{array}\)
Vậy \(x = \dfrac{9}{2}\) hoặc \(x = \dfrac{1}{2}.\)
\(\begin{array}{l}d)\;\sqrt {{{\left( {3x - 2} \right)}^2}} = \left| { - 2} \right|\\ \Leftrightarrow \sqrt {{{\left( {3x - 2} \right)}^2}} = 2\\ \Leftrightarrow \left| {3x - 2} \right| = 2\\ \Leftrightarrow \left[ \begin{array}{l}3x - 2 = 2\\3x - 2 = - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = 4\\3x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{4}{3}\\x = 0\end{array} \right..\end{array}\)
Vậy \(x = \dfrac{4}{3}\) hoặc \(x = 0.\)
\(\begin{array}{l}e)\;\sqrt {{{\left( {x - 2} \right)}^2}} = 2x - 1\\ \Leftrightarrow \left\{ \begin{array}{l}2x - 1 \ge 0\\{\left( {x - 2} \right)^2} = {\left( {2x - 1} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x \ge 1\\{x^2} - 4x + 4 = 4{x^2} - 4x + 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{1}{2}\\3{x^2} = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{1}{2}\\{x^2} = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{1}{2}\\\left[ \begin{array}{l}x = 1\\x = - 1\end{array} \right.\end{array} \right. \Leftrightarrow x = 1.\end{array}\)
Vậy \(x = 1.\)