Question

[๖ۣۜWin๖ۣۜZy (ツтєαм♕¢âи♕6 ᴾᴿᴼシ)]

Tìm x biết

(x+20) ^ 2020 + |y+4|^2019 =0

Answer 1

[Trí Tiên亗]

\(\left(x+20\right)^{2020}+\left|y+4\right|^{2019}=0\)

Ta thấy : \(\hept{\begin{cases}\left(x+20\right)^{2020}\ge0\forall x\\\left|y+4\right|^{2019}\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x+20\right)^{2020}+\left|y+4\right|^{2019}\ge0\forall x,y\)

Do đó, dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+20\right)^{2020}=0\\\left|y+4\right|^{2019}=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

Vậy : \(\left(x,y\right)=\left(-20,-4\right)\)

Answer 2

[chuyên toán thcs ( Cool Team )]

( x + 20 )²⁰²⁰  + | y + 4 |²⁰¹⁹ = 0

Vì ( x + 20 )²⁰²⁰ \(\ge\)0

 | y + 4 |²⁰¹⁹ \(\ge\) 0

=> ( x + 20 )²⁰²⁰  + | y + 4 |²⁰¹⁹ \(\ge\)0

Dấu " = " xảy ra khi 

\(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}}\)

Vậy ................................