Cách giải:
$a,(x+y)^2+|y+\dfrac{1}{4}| \leq 0$
$\begin{cases}(x+y)^2 \geq 0\\|y+\dfrac{1}{4}| \geq 0\\\end{cases}$
$\to (x+y)^2+|y+\dfrac{1}{4}| \geq 0$
$\to (x+y)^2+|y+\dfrac{1}{4}|=0$
$\to \begin{cases}y=-\dfrac{1}{4}\\x=-y=\dfrac{1}{4}\\\end{cases}$
Vậy $\begin{cases}y=-\dfrac{1}{4}\\x=-y=\dfrac{1}{4}\\\end{cases}$
$b,|x-y|+(x+\dfrac{2}{5})^4 \leq 0$
$\begin{cases}|x-y| \geq 0\\(x+\dfrac{2}{5})^4 \geq 0\\\end{cases}$
$\to |x-y|+(x+\dfrac{2}{5})^4 \geq 0$
$\to |x-y|+(x+\dfrac{2}{5})^4=0$
$\to \begin{cases}x=-\dfrac{2}{5}\\x=y=-\dfrac{2}{5}\\\end{cases}$
Vậy $\begin{cases}x=-\dfrac{2}{5}\\y=-\dfrac{2}{5}\\\end{cases}$
