A=x2+2xy+2y2-2x-4y+2
=x2+xy-x+y2+xy-y-x-y+1+y2-2y+1
=(x2+xy-x)+(y2+xy-y)-(x+y-1)+(y2-2y+1)
= x(x+y-1)+y(y+x-1)-(x+y-1)+(y-1)2
=(x+y-1)(x+y-1)+(y-1)2
A=(x+y-1)2+(y-1)2
do (x+y-1)2\(\ge0\forall x;y\)
(y-1)2\(\ge0\forall y\)
=>(x+y-1)2+(y-1)2\(\ge0\)
=>Min A=0 khi
x+y-1=0
=>x+y=1 (*)
y-1=0
=>y=1
thay y=1 vào (*) ta đc
x+1=1
=>x=0
vậy....
3) \(B=3x^2+x+7\)
\(\Leftrightarrow B=3x^2+x+\dfrac{1}{12}+\dfrac{83}{12}\)
\(\Leftrightarrow B=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)
\(\Leftrightarrow B=3\left[x^2+2.x.\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2\right]+\dfrac{83}{12}\)
\(\Leftrightarrow B=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)
Vậy GTNN của \(B=\dfrac{83}{12}\) khi \(x+\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{-1}{6}\)
B=3x2 +x+7
=3x2+x+\(\dfrac{1}{12}+\dfrac{83}{12}\)
=3\(\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)
=3 \(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)
do \(\left(x+\dfrac{1}{6}\right)^2\ge0\forall x\)
=> \(3\left(x+\dfrac{1}{6}\right)^2\ge0\)
=> 3\(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\ge\dfrac{83}{12}\)
GTNN B=\(\dfrac{83}{12}\)
khi x+\(\dfrac{1}{6}\) =0
=>x=-\(\dfrac{1}{6}\)