Question

Tìm GTNN cùa:

a, \(2x^2+y^2+4x-2y-2xy+10\)

b,\(5x^2+y^2+2xy-4x\)

c,\(x^2-4xy+5y^2+10x-22y+28\)

Answer 1

a)\(2x^2+y^2+4x-2y-2xy+10=2x^2+y^2+4x-2y\left(x+1\right)+10\)

\(=y^2-2y\left(x+1\right)+2\left(x^2+2x+1\right)+8\)

\(=y^2-2y\left(x+1\right)+2\left(x+1\right)^2+8\)

\(=\left(y+x+1\right)^2+\left(x+1\right)^2+8\ge8\)

Dấu "=" xảy ra khi x=-1 và y=0

Answer 2

b)\(5x^2+y^2+2xy-4x=\left(x^2+2xy+y^2\right)+\left(4x^2-4x+1\right)-1\)

\(=\left(x+y\right)^2+\left(2x-1\right)^2-1\ge-1\)

Dấu "=" xảy ra khi x=1/2 và y=-1/2

Answer 3

c)\(x^2-4xy+5y^2+10x-22y+28=x^2-2x\left(2y-5\right)+5y^2-22y+28\)

\(=x^2-2x\left(2y-5\right)+\left(4y^2-20y+25\right)+\left(y^2-2y+1\right)+2\)

\(=x^2-2x\left(2y-5\right)+\left(2y-5\right)^2+\left(y-1\right)^2+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu "=" xảy ra khi x=-3 và y=1