Đề bài
Giải các phương trình sau:
a) \(2{x^2} + 3x = 0\)
b) \({x^2} - 2x - 15 = 0\)
c) \(4{x^2} - 9 = 0\)
Lời giải chi tiết
\(\begin{array}{l}a)\;2{x^2} + 3x = 0\\ \Leftrightarrow x\left( {2x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\2x + 3 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = - \dfrac{3}{2}\end{array} \right.\end{array}\)
\(\begin{array}{l}b)\;{x^2} - 2x - 15 = 0\\ \Leftrightarrow {x^2} - 5x + 3x - 15 = 0\\ \Leftrightarrow x\left( {x - 5} \right) + 3\left( {x - 5} \right) = 0\\ \Leftrightarrow \left( {x - 5} \right)\left( {x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\x + 3 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = - 3\end{array} \right.\end{array}\)
\(\begin{array}{l}c)\;4{x^2} - 9 = 0\\ \Leftrightarrow {\left( {2x} \right)^2} - {3^2} = 0\\ \Leftrightarrow \left( {2x - 3} \right).\left( {2x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2x - 3 = 0\\2x + 3 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{3}{2}\\x = - \dfrac{3}{2}\end{array} \right.\end{array}\)
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