a)9/4:9/2.|1/2-1/2x^2|=3/4
⇔ $\frac{1}{2}$.|$\frac{1}{2}$ - $\frac{1}{2x²}$| = $\frac{3}{4}$
⇔ |$\frac{1}{2}$ - $\frac{1}{2x²}$| = $\frac{3}{2}$
⇔ \(\left[ \begin{array}{l}\frac{1}{2}-\frac{1}{2x²}=\frac{3}{2}\\\frac{1}{2}-\frac{1}{2x²}=-\frac{3}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\frac{1}{2x²}=-1\\\frac{1}{2x²}=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x²=-\frac{1}{2}(vô lí)\\x²=\frac{1}{4}\end{array} \right.\)
⇒ x = ±$\frac{1}{2}$
c) (x+2y)² + |x-2/5|=0
Vì (x + 2y)² ≥ 0 và |x - $\frac{2}{5}| ≥ 0 mà (x+2y)² + |x-$\frac{2}{5}$|=0
⇒ $\left \{ {{|x-\frac{2}{5}|=0} \atop {(x+2y)²=0}} \right.$
⇔ $\left \{ {{x-\frac{2}{5}=0} \atop {x+2y=0}} \right.$
⇔ $\left \{ {{x=\frac{2}{5}} \atop {x+2y=0}} \right.$
⇔ $\left \{ {{x=\frac{2}{5}} \atop {\frac{2}{5}+2y=0}} \right.$
⇔ $\left \{ {{x=\frac{2}{5}} \atop {2y=-\frac{2}{5}}} \right.$
⇔ $\left \{ {{x=\frac{2}{5}} \atop {y=-\frac{1}{5}}} \right.$
