\(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2020}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall xy.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\ge0\) \(\forall xy.\)
Mà \(\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}\le0.\)
\(\Rightarrow\left(2x-5\right)^{2020}+\left(3y+4\right)^{2018}=0\)
\(\Rightarrow\left(2x-5\right)+\left(3y+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
Ta có:
(2x+5)2020 ≥ 0 với ∀ x
(3y+4)2018 ≥ 0 với ∀ y
⇒ (2x+5)2020 + (3y+4)2018 ≥ 0 với ∀ x, y
Mà (2x+5)2020 + (3y+4)2018 ≤ 0
⇒ (2x+5)2020 + (3y+4)2018 = 0
⇒ \(\left[{}\begin{matrix}\left(2x+5\right)^{2020}=0\\\left(3y+4\right)^{2018}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x+5=0\\3y+4=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}2x=-5\\3y=-4\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{-5}{2}\\y=\frac{-4}{3}\end{matrix}\right.\)
Vậy...
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