a: \(M=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)

\(=-2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}\right)\)

\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\forall x\)

Dấu '=' xảy ra khi x=3/4

b: Tham khảo:

\(A=-\left(x^2-2x\left(y+1\right)+\left(y+1\right)^2\right)-\left(4y^2-10y-5-\left(y+1\right)^2\right)\)

\(=-\left(x-y-1\right)^2-\left(3y^2-12y-6\right)\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+18\le18\)

Max A=18 khi y=2; x=3

\(B=-\left(x^2+2x\left(y-1\right)+\left(y-1\right)^2\right)-\left(2y^2+2y-\left(y-1\right)^2\right)-15\)

\(=-\left(x+y-1\right)^2-\left(y+2\right)^2-10\le-10\)

Max B=-10 khi y=-2; x= 3

Đặt A=\(-x^2+2x\left(y+1\right)-\left(y-1\right)^2-3y^2+8y+6\)

=\(-\left(x-y+1\right)^2-3\left(y^2-\frac{8}{3}y+\frac{16}{9}\right)+\frac{34}{3}\)

=\(-\left(x-y+1\right)^2-3\left(y-\frac{4}{3}\right)^2+\frac{34}{3}\le\frac{34}{3}\)

dấu = xảy ra khi \(\left\{\begin{matrix}x-y+1=0\\y-\frac{4}{3}=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=\frac{1}{3}\\y=\frac{4}{3}\end{matrix}\right.\)

Vậy max A=\(\frac{34}{3}\)khi và chỉ khi x=1/3, y=4/3