a) Sửa: C=(x+2)2+\(\left(y-\frac{1}{5}\right)^2\)+10
Ta có: \(\hept{\begin{cases}\left(x+2\right)^2\ge0\forall x\\\left(y-\frac{1}{5}\right)^2\ge0\forall y\end{cases}}\)
\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2+10\ge10\forall x;y\)
hay C \(\ge10\). Dấu "=" \(\Leftrightarrow\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-\frac{1}{5}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{5}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}}}\)
@Quỳnh : Sao phải sửa ?
a) \(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\)
Ta thấy : \(\left(x+2\right)^2\ge0,\forall x\)
\(\left(y-\frac{1}{5}\right)^2\ge0,\forall y\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge-10\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x+2=0\\y-\frac{1}{5}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2\\y=\frac{1}{5}\end{cases}}\)
Vậy \(Min_C=-10\Leftrightarrow\left(x;y\right)=\left(-2;\frac{1}{5}\right)\)
b) Để D max
\(\Leftrightarrow\left(2x-3\right)^2+5\)min
Ta có : \(\left(2x-3\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(2x-3\right)^2+5\ge5\)
Dấu " = " xảy ra :
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy \(Max_D=\frac{4}{5}\Leftrightarrow x=\frac{3}{2}\)
