Giải các phương trình:
a) \(\frac{{\sqrt {4{x^2} + 7x - 2} }}{{x + 2}} = \sqrt 2 \)
b) \(\sqrt {2{x^2} + 3x - 4} = \sqrt {7x + 2} \)
a) ĐKXĐ: \(\left\{ \begin{array}{l}
4{x^2} + 7x - 2 \ge 0\\
x \ne - 2
\end{array} \right.\)
\(\frac{{\sqrt {4{x^2} + 7x - 2} }}{{x + 2}} = \sqrt 2 \)
\(\begin{array}{l}
\Rightarrow 4{x^2} + 7x - 2 = 2{\left( {x + 2} \right)^2}\\
\Leftrightarrow 2{x^2} - x - 10 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{5}{2}\,\,\,\left( n \right)\\
x = - 2\,\,\left( l \right)
\end{array} \right.
\end{array}\)
Vậy phương trình có 1 nghiệm là \(x = \frac{5}{2}\)
b) \(\sqrt {2{x^2} + 3x - 4} = \sqrt {7x + 2} \)
\(\begin{array}{l}
\sqrt {2{x^2} + 3x - 4} = \sqrt {7x + 2} \\
\Leftrightarrow \left\{ \begin{array}{l}
7x + 2 \ge 0\\
2{x^2} + 3x - 4 = 7x + 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{2}{7}\\
2{x^2} - 4x - 6 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{2}{7}\\
x = 3\left( n \right) \vee x = - 1\left( l \right)
\end{array} \right. \Leftrightarrow x = 3
\end{array}\)
-- Mod Toán 10