Bài 3 trang 19 Tài liệu dạy – học Toán 9 tập 1

Đề bài

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) \(\sqrt {{{\left( {x - 2} \right)}^2}}  = 2x - 1\).

Lời giải chi tiết

\(a)\;\;\sqrt {2x - 1}  = 5\)

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 \(x = 1.\)