b: \(P=-2\left(x^2-3x-\dfrac{9}{2}\right)\)
\(=-2\left(x^2-3x+\dfrac{9}{4}-\dfrac{27}{4}\right)\)
\(=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{2}\le\dfrac{27}{2}\)
Dấu '=' xảy ra khi x=3/2
c: \(C=-4\left(x^2+\dfrac{5}{4}x-\dfrac{1}{4}\right)\)
\(=-4\left(x^2+2\cdot x\cdot\dfrac{5}{8}+\dfrac{25}{64}-\dfrac{41}{64}\right)\)
\(=-4\left(x+\dfrac{5}{8}\right)^2+\dfrac{41}{16}\le\dfrac{41}{16}\)
Dấu '=' xảy ra khi x=-5/8
\(D=x^2+2x\left(y+2\right)+2y^2+6y+10\)
\(D=x^2+2x\left(y+2\right)+\left(y^2+4y+4\right)+y^2+2y+6\)
\(D=\left[x+\left(y+2\right)\right]^2+\left(y+1\right)^2+5\ge5\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+\left(y+2\right)=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+-1+2=0\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\)
Vậy \(Min_D=5\) khi \(\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\)
