Tìm x,y: a, ( x + y )^2020 + I 2021 – y I ≤ 0 b, I 3x + 2y I^209 + I 4y – 1 I^2020 ≤ 0 09/11/2021 Bởi Melody Tìm x,y: a, ( x + y )^2020 + I 2021 – y I ≤ 0 b, I 3x + 2y I^209 + I 4y – 1 I^2020 ≤ 0
a, Ta có: $\begin{cases}(x+y)^{2020}\ge0 \ ∀x;y\\|2021-y|\ge0 \ ∀y\end{cases}$ $\to (x+y)^{2020}+|2021-y|\ge0 \ ∀x;y$ Mà $(x+y)^{2020}+|2021-y|\le0$ $\to (x+y)^{2020}+|2021-y|=0$ $↔\begin{cases}x+y=0\\2021-y=0\end{cases}↔\begin{cases}x=-2021\\y=2021\end{cases}$ Vậy $(x;y)=(-2021;2021)$ b, Ta có: $\begin{cases}|3x+2y|^{209}\ge0 \ ∀x;y\\|4y-1|^{2020}\ge0 \ ∀y\end{cases}$ $\to |3x+2y|^{209}+|4y-1|^{2020}\ge0$ Mà $|3x+2y|^{209}+|4y-1|^{2020}\le0$ $\to |3x+2y|^{209}+|4y-1|^{2020}=0$ $↔\begin{cases}3x+2y=0\\4y-1=0\end{cases}↔\begin{cases}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{cases}$ Vậy `(x;y)=(-1/6;1/4)` Bình luận
Tham khảo `a)` Vì `(x+y)^{2020}≥0∀x;y` `|2021-y|≥0∀y` Mà `(x+y)^{2020}+|2021-y|≤0` `→`\(\left[ \begin{array}{l}x+y=0\\2021-y=0\end{array} \right.\) `→`\(\left[ \begin{array}{l}x=-y\\y=2021\end{array} \right.\) `→`\(\left[ \begin{array}{l}x=-2021\\y=2021\end{array} \right.\) Vậy `x=-2021;y=2021` `b)` Vì `|3x+2y|^{209}≥0∀x;y` `|4y-1|^{2020}≥0∀y` Mà `|3x+2y|^{209}+|4y-1|^{2020}≤0` `→`\(\left[ \begin{array}{l}3x+2y=0\\4y-1=0\end{array} \right.\) `→`\(\left[ \begin{array}{l}3x=-2y\\4y=1\end{array} \right.\) `→`\(\left[ \begin{array}{l}3x=-2y\\y=\dfrac{1}{4}\end{array} \right.\) `→`\(\left[ \begin{array}{l}3x=\dfrac{-1}{2}\\y=\dfrac{1}{4}\end{array} \right.\) `→`\(\left[ \begin{array}{l}x=\dfrac{-1}{6}\\y=\dfrac{1}{4}\end{array} \right.\) Vậy ` x=\frac{-1}{6};y=\frac{1}{4}` Bình luận
a,
Ta có:
$\begin{cases}(x+y)^{2020}\ge0 \ ∀x;y\\|2021-y|\ge0 \ ∀y\end{cases}$
$\to (x+y)^{2020}+|2021-y|\ge0 \ ∀x;y$
Mà $(x+y)^{2020}+|2021-y|\le0$
$\to (x+y)^{2020}+|2021-y|=0$
$↔\begin{cases}x+y=0\\2021-y=0\end{cases}↔\begin{cases}x=-2021\\y=2021\end{cases}$
Vậy $(x;y)=(-2021;2021)$
b,
Ta có:
$\begin{cases}|3x+2y|^{209}\ge0 \ ∀x;y\\|4y-1|^{2020}\ge0 \ ∀y\end{cases}$
$\to |3x+2y|^{209}+|4y-1|^{2020}\ge0$
Mà $|3x+2y|^{209}+|4y-1|^{2020}\le0$
$\to |3x+2y|^{209}+|4y-1|^{2020}=0$
$↔\begin{cases}3x+2y=0\\4y-1=0\end{cases}↔\begin{cases}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{cases}$
Vậy `(x;y)=(-1/6;1/4)`
Tham khảo
`a)` Vì `(x+y)^{2020}≥0∀x;y`
`|2021-y|≥0∀y`
Mà `(x+y)^{2020}+|2021-y|≤0`
`→`\(\left[ \begin{array}{l}x+y=0\\2021-y=0\end{array} \right.\)
`→`\(\left[ \begin{array}{l}x=-y\\y=2021\end{array} \right.\)
`→`\(\left[ \begin{array}{l}x=-2021\\y=2021\end{array} \right.\)
Vậy `x=-2021;y=2021`
`b)` Vì `|3x+2y|^{209}≥0∀x;y`
`|4y-1|^{2020}≥0∀y`
Mà `|3x+2y|^{209}+|4y-1|^{2020}≤0`
`→`\(\left[ \begin{array}{l}3x+2y=0\\4y-1=0\end{array} \right.\)
`→`\(\left[ \begin{array}{l}3x=-2y\\4y=1\end{array} \right.\)
`→`\(\left[ \begin{array}{l}3x=-2y\\y=\dfrac{1}{4}\end{array} \right.\)
`→`\(\left[ \begin{array}{l}3x=\dfrac{-1}{2}\\y=\dfrac{1}{4}\end{array} \right.\)
`→`\(\left[ \begin{array}{l}x=\dfrac{-1}{6}\\y=\dfrac{1}{4}\end{array} \right.\)
Vậy ` x=\frac{-1}{6};y=\frac{1}{4}`