\(A=\left(x^2+3x+4\right)^2\)

ta có:

\(x^2+3x+4=x^2+2\cdot\dfrac{3}{2}x+\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}\\ =\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

vậy \(minA=\left(\dfrac{7}{4}\right)^2=\dfrac{49}{16}\Leftrightarrow x=-\dfrac{3}{2}\)

b: Ta có: \(B=-x^2-y^2+2x-6y+9\)

\(=-\left(x^2-2x+y^2+6y-9\right)\)

\(=-\left(x^2-2x+1+y^2+6y+9-19\right)\)

\(=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=-3

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

hay \(x\in\left\{3;1;4;0;9;-5;16;-12\right\}\)

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)