Question

tìm giá trị nhỏ nhất: 8x^2 + 2y^2 +4xy —2x +4y +2015

Answer 1

\(P=8x^2+2y^2+4xy-2x+4y+2015=2\cdot\left(y^2+2xy+2y+4x^2-x\right)+2015\)

\(=2\cdot\left(y^2+2y\left(x+1\right)+\left(x+1\right)^2-\left(x+1\right)^2+4x^2-x\right)+2015\)

\(=2\cdot\left[\left(y+\left(x+1\right)\right)^2+3x^2-3x-1\right]+2015\)

\(=2\cdot\left[\left(y+x+1\right)^2+3\left(x^2-2x\cdot\frac{1}{2}+\frac{1}{4}\right)-1-\frac{3}{4}\right]+2015\)

\(=2\cdot\left[\left(y+x+1\right)^2+3\cdot\left(x-\frac{1}{2}\right)^2\right]+2015-\frac{7}{2}\)

\(=2\cdot\left(x+y+1\right)^2+6\left(x-\frac{1}{2}\right)^2+2011\frac{1}{2}\)

Vậy GTNN của P = 2011,5. Xảy ra khi x=0,5 và y=-1,5.