Question

tìm các số nguyên x,y thỏa mãn( x+3)²⁰²⁰+(y-2)²⁰²⁰=0

Answer 1

\(\left(x+3\right)^{2020}+\left(y-2\right)^{2020}=0\)

Vì \(\left(x+3\right)^{2020}\ge0\forall x;\left(y-2\right)^{2020}\ge0\forall y\)

\(\Rightarrow\left(x+3\right)^{2020}+\left(y-2\right)^{2020}\ge0\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+3\right)^{2020}=0\\\left(y-2\right)^{2020}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+3=0\\y-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}}\)

Vậy ....

Answer 2

(x−6)2020+2(y−3)2020=0

Ta có : (�−6)2020≥0∀�(x−6)2020≥0∀x

            2(�+3)2020≥0∀�2(y+3)2020≥0∀y

        =>(�−6)2020+2(�+3)2020≥0∀�,�(x−6)2020+2(y+3)2020≥0∀x,y

Dấu "=" xảy ra <=>{�−6=0�+3=0<=>{�=6�=−3{x−6=0y+3=0​<=>{x=6y=−3​