Question

Tìm x và y: (x-3,5)^2+(y-1/10)^4≤0

Answer 1

\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)

Vì: \(\left(x-3,5\right)^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-3,5\right)^2=0\\\left(y-\dfrac{1}{10}\right)^4=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

Answer 2

Ta có: \(\left(x-3.5\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)

Do đó: \(\left(x-\dfrac{7}{2}\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{7}{2};\dfrac{1}{10}\right)\)