Question

[Bảo Nguyễn Gia]

A=x\(^{2024}\)+(5-y)\(^3\) biết |x+1|+(y-2)\(^{2028}\)=0

Answer 1

[Phạm Trần Hoàng Anh]

Ta có: \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left(y-2\right)^{2028}\ge0\end{matrix}\right.\) 

=> \(\left|x+1\right|+\left(y-2\right)^{2028}\ge0\)

Dấu = xảy ra khi: 

\(\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Khi đó: 

`A =` \(x^{2024}+\left(5-y\right)^3\)

`=` \(\left(-1\right)^{2024}\) `+ (5-2)^3 `

`= 1 + 3^3 `

`=1 + 27`

`= 28`

Vậy `A = 28`

Answer 2

[Bảo Nguyễn Gia]

giúp mình với mình dg vội