Tìm x, biết :
a) \(\sqrt {{x^2}} = 5\);
b) \(\sqrt {{x^2}} = \left| { - 2} \right|\);
c) \(\sqrt {16{x^2}} = 3\);
d) \(\sqrt {9{x^2}} = \left| { - 2} \right|\).
Advertisements (Quảng cáo)
Phương trình \(\sqrt {{x^2}} = A \Leftrightarrow \left| x \right| = A \Leftrightarrow \left[ \begin{array}{l}x = A\\x = - A\end{array} \right..\)
\(\begin{array}{l}a)\;\;\sqrt {{x^2}} = 5\\ \Leftrightarrow \left| x \right| = 5\\ \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 5\end{array} \right..\end{array}\)
\(\begin{array}{l}b)\;\sqrt {{x^2}} = \left| { - 2} \right|\\ \Leftrightarrow \left| x \right| = 2\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\\x = - 2\end{array} \right..\end{array}\)
\(\begin{array}{l}c)\;\sqrt {16{x^2}} = 3\\ \Leftrightarrow \left| {4x} \right| = 3\\ \Leftrightarrow \left[ \begin{array}{l}4x = 3\\4x = - 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{3}{4}\\x = - \dfrac{3}{4}\end{array} \right..\end{array}\)
\(\begin{array}{l}d)\;\sqrt {9{x^2}} = \left| { - 2} \right|\\ \Leftrightarrow \left| {3x} \right| = 2\\ \Leftrightarrow \left[ \begin{array}{l}3x = 2\\3x = - 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{2}{3}\\x = - \dfrac{2}{3}\end{array} \right..\end{array}\)