Tìm các giá trị x, y thỏa mãn: $|2x-27|^{2021}+(3y+10)^{2020}=0$

Vì $\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.$

⇒ $\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y$

Dấu "=" ⇔ $\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.$

⇒ $\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.$

Vậy ...............

Answer

Ta có:

`|2x - 27|^{2021} >= 0 AA x`

`(3y + 10)^{2020} >= 0 AA y`

`=> |2x - 27|^{2021} + (3y + 10)^{2020} >= 0 AA x , y`

Dấu $"="$ xảy ra

`<=>` `{(2x - 27 = 0),(3y + 10 = 0):}`

`<=>` `{(2x = 0 + 27),(3y = 0 - 10):}`

`<=>` `{(2x = 27),(3y = -10):}`

`<=>` `{(x = 27 : 2),(y = -10 : 3):}`

`<=>` $\begin{cases} x = \dfrac{27}{2}\\y = \dfrac{-10}{3} \end{cases}$

Vậy `(x , y) = (27/2 ; {-10}/3)`