Question

Tìm GTNN

\(B=\frac{3y^2}{-25x^2+20xy-5y^2}\)

Answer 1

B= \(\frac{3y^2}{-25x^2+20xy-5y^2}=\frac{3y^2}{-y^2-\left(25x^2-20xy+4y^2\right)}=\frac{1}{-\frac{y^2}{3y^2}-\frac{\left(5x-2y\right)^2}{3y^2}}\)

=\(\frac{1}{-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}}\)

Có \(\frac{1}{3}+\frac{\left(5x-2y\right)^2}{3y^2}\ge\frac{1}{3}\) vs mọi x,y và y\(\ne0\)

<=>\(-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}\le-\frac{1}{3}\)

<=> \(\frac{1}{-\frac{1}{3}-\frac{\left(5x-2y\right)^2}{3y^2}}\ge-3\) <=> B \(\ge3\)

Dấu "=" xảy ra <=> 5x-2y=0

<=> 5x=2y < => \(x=\frac{2y}{5}\)

Vậy minB=3 <=> \(x=\frac{2y}{5}\)