Câu 1 : x2+3x+3 = (x2 + 2.\(\frac{3}{2}\).x + \(\frac{9}{4}\)) - \(\frac{9}{4}\)+ 3
= (x2 + 2.\(\frac{3}{2}\).x + \(\frac{9}{4}\)) + \(\frac{3}{4}\)= ( x+ \(\frac{3}{2}\))2 + \(\frac{3}{4}\)
Ta có:( x+ \(\frac{3}{2}\))2 ≥ 0 vs mọi x
<=>( x+ \(\frac{3}{2}\))2 + \(\frac{3}{4}\) ≥ \(\frac{3}{4}\)
Dấu '' ='' xãy ra <=> x + \(\frac{3}{2}\)=0
=> x =-\(\frac{3}{2}\)
Vậy vs x =-\(\frac{3}{2}\)thì min A = \(\frac{3}{4}\)
Bài 2:
Đặt \(A=2x-2xy-2x^2-y^2\)
\(-A=2x^2+y^2+2xy-2x=(x^2+y^2+2xy)+(x^2-2x)\)
\(=(x+y)^2+(x^2-2x+1)-1=(x+y)^2+(x-1)^2-1\)
Ta thấy:
$(x+y)^2\geq 0; (x-1)^2\geq 0$ với mọi $x,y$
$\Rightarrow -A=(x+y)^2+(x-1)^2-1\geq -1$
$\Rightarrow A\leq 1$
Vậy $A_{\max}=1$
Dấu "=" xảy ra khi \(\left\{\begin{matrix} x+y=0\\ x-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=-1\end{matrix}\right.\)