Tìm x,y:a, ( x + y )^2020 + I 2021 - y I ≤ 0b, I 3x + 2y I^209 + I 4y - 1 I^2020 ≤ 0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\left(x+y\right)^{2020}+\left|2021-y\right|\le0\)
Dấu ''='' xảy ra \(\Leftrightarrow\hept{\begin{cases}x=-y\\y=2021\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2021\\y=2021\end{cases}}}\)
b, \(\left|3x+2y\right|^{209}+\left|4y-1\right|^{2020}\le0\)
Dấu ''='' xảy ra <=> \(\hept{\begin{cases}3x=-2y\\4y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-2y\\y=\frac{1.}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-\frac{1}{2}\\y=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{6}\\y=\frac{1}{4}\end{cases}}\)Vậy \(\left\{x;y\right\}=\left\{-\frac{1}{6};\frac{1}{4}\right\}\)
a: \(\Leftrightarrow\left(2x-1;y-3\right)\in\left\{\left(1;10\right);\left(5;2\right);\left(-1;-10\right);\left(-5;-2\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;13\right);\left(3;5\right)\right\}\)
b: \(\Leftrightarrow\left(3x-2;2y-3\right)\in\left\{\left(-1;-1\right);\left(1;1\right)\right\}\)
hay \(\left(x,y\right)\in\left(1;2\right)\)
c: \(\Leftrightarrow\left(x+1,2y-1\right)\in\left\{\left(12;1\right);\left(4;3\right);\left(-12;-1\right);\left(-4;-3\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(11;1\right);\left(3;2\right)\right\}\)
`|x-1|+2020|x-2|+|x-3|`
`=|x-1|+|3-x|+2020|x-2|`
Áp dụng BĐT `|A|+|B|>=|A+B|`
`=>|x-1|+|3-x|>=|x-1+3-x|=2`
Mà `|x-2|>=0=>2020|x-2|>=0`
`=>|x-1|+2020|x-2|+|x-3|>=2`
Dấu "=" xảy ra khi $\begin{cases}(x-1)(3-x) \ge 0\\x-2=0\\\end{cases}$
`<=>` $\begin{cases}(x-1)(x-3) \le 0\\x=2\\\end{cases}$
`<=>` $\begin{cases}1 \le x \le 3\\x=2\\\end{cases}$
`<=>x=2`