Question

( x - 2)²⁰¹² + |y² - 9|²⁰¹⁴ = 0

Answer 1

Ta có \(\hept{\begin{cases}\left(x-2\right)^{2012}\ge0\\\left|y^2-9\right|^{2014}\ge0\end{cases}\forall x,y}\)

\(\Leftrightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}\ge0\forall x,y\)

Do đó để ( x - 2)²⁰¹² + |y² - 9|²⁰¹⁴ = 0 \(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\\left|y^2-9\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y^2-9=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y^2=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)  hoặc \(\hept{\begin{cases}x=2\\y=-3\end{cases}}\)

Vậy \(\hept{\begin{cases}x=2\\y=-3\end{cases}}\)  hoặc \(\hept{\begin{cases}x=2\\y=3\end{cases}}\)

~~~~ Học tốt ~~~~~

Answer 2

Vì \(\left(x-2\right)^{2012}\ge0\forall x\)và \(\left|y^2-9\right|^{2014}\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}\ge0\forall x,y\)

mà \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)( giả thiết )

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y^2-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y^2=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\pm3\end{cases}}\)

Vậy \(\hept{\begin{cases}x=2\\y=\pm3\end{cases}}\)