\(B=2x-x^2-5\)
\(-B=x^2-2x+5\)
\(-B=\left(x^2-2x+1\right)+4\)
\(-B=\left(x-1\right)^2+4\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow-B\ge4\)
\(\Leftrightarrow B\le-4\)
Dấu " = " xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy ...
\(K=-x^2-y^2-x+6y+10\)
\(-K=x^2+y^2+x-6y-10\)
\(-K=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-6y+9\right)-\frac{77}{4}\)
\(-K=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{77}{4}\)
Mà \(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\left(y-3\right)^2\ge0\)
\(\Rightarrow-K\ge-\frac{77}{4}\)
\(\Leftrightarrow K\le\frac{77}{4}\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x+\frac{1}{2}=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)
Vậy ...
