Question

tìm giá trị nhỏ nhất

A=x^2+y^2-x+4y+5

B=2x^2+4y^2+4xy-3x-1

thanks trc ạ

Answer 1

\(x^2+y^2-x+4y+5\)

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

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

\(\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(x=\frac{1}{2};y=-2\)

Answer 2

\(B=2x^2+4y^2+4xy-3x-1\)

\(=\left(x^2+4xy+4y^2\right)+\left(x^2-3x+\frac{9}{4}\right)-\frac{13}{4}\)

\(=\left(x+2y\right)^2+\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\)

\(\ge-\frac{13}{4}\)

Dấu "=" xảy ra khi \(x=\frac{3}{2};y=-\frac{3}{4}\)