Question

Tìm x,y biết :

(2x-1)²⁰⁰⁸+(y+3.1)²⁰⁰⁸=0

Answer 1

(2x-1)²⁰⁰⁸+(y+3.1)²⁰⁰⁸=0

ĐK: \(\hept{\begin{cases}\left(2x-1\right)^{2008}\ge0\\\left(y+3.1\right)^{2008}\ge0\end{cases}}\Rightarrow\left(2x-1\right)^{2008}+\left(y+3\right)^{2008}\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x-1\right)^{2008}=0\\\left(y+3\right)^{2008}=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-1=0\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

Vậy x=1/2 và y=-3