\(A=\)\(x^2+y^2-4x+y+5.\)
\(=\left(x^2-4x+4\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-2\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow A_{min}=\frac{3}{4}\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y+\frac{1}{2}\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}}}\)
\(x^2+y^2-4x+y+5=\left(x-2\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow Min=\frac{3}{4}\)Dấu "=" xr \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}}}\)
\(=x^2-4y+4+y^2+y+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-2\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x-2\right)^2\ge0\forall x\in R\)
\(\left(y+\frac{1}{2}\right)^2\ge0\forall y\in R\)
\(\Rightarrow\left(x-2\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> GTNN của biểu thức =\(\frac{3}{4}\) tại x=2 và y =\(\frac{-1}{2}\)
Vậy..........
hc tốt
