Ta có:
\(R=x^2+y^2-x+6y+10\)
\(R=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}+y^2+3y+3y+9+\dfrac{3}{4}\)
\(R=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Hay \(R\ge\dfrac{3}{4}\) với mọi giá trị của \(x;y\in R\).
Để \(R=\dfrac{3}{4}\) thì \(\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
\Vậy=..
Chúc bạn học tốt!!!
