a, A=x(x-6)+10
=x2-6x+10=(x2-2.3.x+9)+10-9
=(x-3)2+1
Ta có : \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0với\forall x\\1>0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+1>0với\forall x\)
Vậy A>0 với \(\forall x\)
b, B=x2-2x+9y2-6y+3
= (x2-2x)+(9y2-6y)+3
=(x2-2.\(\frac{1}{2}x+\frac{1}{4}\))+(9y2-2.3y.1+1)+3-1-\(\frac{1}{4}\)
=(x-\(\frac{1}{4}\))2+\(\left(3y-1\right)^2+\frac{7}{4}\)
ta có : \(\left\{{}\begin{matrix}\left(x-\frac{1}{4}\right)^2\ge0với\forall x\\\left(3y-1\right)^2\ge0với\forall x\\\frac{7}{4}>0\end{matrix}\right.\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2+\left(3y-1\right)^2+\frac{7}{4}>0với\forall x\)
Vậy B>0 với \(\forall x\)
