Câu hỏi
Giải các phương trình sau:
a) \(\sqrt {4{x^2}} = 6\)
b) \(\sqrt {{x^2}} = \left| { - 8} \right|\)
c) \(\sqrt {{x^2} - 22x + 121} = 2x - 15\)
d) \(\sqrt {1 + 9{x^2} - 6x} = 2x + 6\)
- A \(\begin{array}{l}
a)\,\,\,S = \left\{ { - 3;\,\,3} \right\}\\
b)\,\,S = \left\{ {8} \right\}\\
c)\,\,S = \left\{ {\frac{{26}}{3}} \right\}\\
d)\,\,S = \left\{ { - 1;\,\,7} \right\}
\end{array}\) - B \(\begin{array}{l}
a)\,\,\,S = \left\{ { - 3;\,\,3} \right\}\\
b)\,\,S = \left\{ {8;\,\, - 8} \right\}\\
c)\,\,S = \left\{ {-\frac{{26}}{3}} \right\}\\
d)\,\,S = \left\{ { - 1;\,\,7} \right\}
\end{array}\) - C \(\begin{array}{l}
a)\,\,\,S = \left\{ { - 3;\,\,3} \right\}\\
b)\,\,S = \left\{ {8;\,\, - 8} \right\}\\
c)\,\,S = \left\{ {\frac{{26}}{3}} \right\}\\
d)\,\,S = \left\{ { - 1;\,\,7} \right\}
\end{array}\) - D \(\begin{array}{l}
a)\,\,\,S = \left\{ {3} \right\}\\
b)\,\,S = \left\{ {8;\,\, - 8} \right\}\\
c)\,\,S = \left\{ {\frac{{26}}{3}} \right\}\\
d)\,\,S = \left\{ { - 1;\,\,7} \right\}
\end{array}\)
Lời giải chi tiết:
Giải các phương trình sau:
a) \(\sqrt {4{x^2}} = 6 \Leftrightarrow \left| {2x} \right| = 6 \Leftrightarrow \left[ \begin{array}{l}2x = 6\\2x = - 6\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = - 3\end{array} \right..\)
Vậy phương trình có hai nghiệm phân biệt \(x = 3,\,\,x = - 3.\)
b) \(\sqrt {{x^2}} = \left| { - 8} \right| \Leftrightarrow \left| x \right| = 8 \Leftrightarrow \left[ \begin{array}{l}x = 8\\x = - 8\end{array} \right..\)
Vậy phương trình có hai nghiệm phân biệt \(x = 8,\,\,x = - 8.\)
c) \(\sqrt {{x^2} - 22x + 121} = 2x - 15\)
\(\begin{array}{l} \Leftrightarrow \sqrt {{{\left( {x - 11} \right)}^2}} = 2x - 15\\ \Leftrightarrow \left\{ \begin{array}{l}2x - 15 \ge 0\\{\left( {x - 11} \right)^2} = {\left( {2x - 15} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{{15}}{2}\\{x^2} - 22x + 121 = 4{x^2} - 60x + 225\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{{15}}{2}\\3{x^2} - 38x + 104 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge \frac{{15}}{2}\\\left[ \begin{array}{l}x = 4\,\,\left( {ktm} \right)\\x = \frac{{26}}{3}\,\,\left( {tm} \right)\end{array} \right.\end{array} \right. \Leftrightarrow x = \frac{{26}}{3}.\end{array}\)
Vậy phương trình có nghiệm duy nhất \(x = \frac{{26}}{3}.\)
d) \(\sqrt {1 + 9{x^2} - 6x} = 2x + 6 \Leftrightarrow \sqrt {{{\left( {3x - 1} \right)}^2}} = 2x + 6\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}2x + 6 \ge 0\\{\left( {3x - 1} \right)^2} = {\left( {2x + 6} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 3\\9{x^2} - 6x + 1 = 4{x^2} + 24x + 36\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge - 2\\5{x^2} - 30x - 35 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 2\\\left( {x + 1} \right)\left( {x - 7} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ge - 2\\\left[ \begin{array}{l}x = - 1\\x = 7\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = 7\end{array} \right..\end{array}\)
Vậy phương trình có hai nghiệm phân biệt \(x = - 1;\,\,x = 7.\)
