Đáp án: $a$) `(x;y)=(-2;-3)`.
$b$) `(x;y)=(0;1)`.
Giải thích các bước giải:
$a$) $(2x+4)^{220} + |xy-6|^{100} ≤ 0$
Vì $(2x+4)^{220} ; |xy-6|^{100} ≥ 0 ∀ x;y$
$⇒(2x+4)^{220} + |xy-6|^{100} ≤ 0$ khi $(2x+4)^{220} + |xy-6|^{100}= 0$
$⇒$ $\left\{\begin{matrix}(2x+4)^{220} = 0& \\|xy-6|^{100}=0& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix}2x+4=0& \\xy-6=0& \end{matrix}\right.$
$⇔ \left\{\begin{matrix}x=-2& \\xy=6& \end{matrix}\right.$
$⇔ \left\{\begin{matrix}x=-2& \\y=-3& \end{matrix}\right.$
Vậy `(x;y)=(-2;-3)`.
$b$) $x^2 + |y-1| ≤ 0$
Vì $x^2; |y-1| ≥ 0 ∀ x;y$
$⇒x^2 + |y-1| ≤ 0$ khi $x^2 + |y-1| = 0$
$⇒$ $\left\{\begin{matrix}x^2=0& \\|y-1|=0& \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix}x=0& \\y=1& \end{matrix}\right.$
Vậy `(x;y)=(0;1)`.
