a) `A=16x^2+y^2-8x+7=(16x^2-8x+1)+(y^2+6)`
`= (4x-1)^2+(y^2+6)`
Vì `(4x-1)^2 ≥0 ; y^2 + 6 ≥ 6>0 \forall x;y`
`=> A_(min) = 6 <=>` \(\left[ \begin{array}{l}4x-1=0\\y^2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{4}\\y=0\end{array} \right.\)
Vậy `A_(min)=6 <=> x=1/4 ; y=0`
b) `B=3x^3-3x^2+x+a \vdots (x-1)`
Có: `x-1 =0 <=> x=1`
Thay `x=1` vào `B` được: `B=1+a`
` => (1+a) \vdots (x-1) <=> 1+a=0 <=> a=-1`