Ta có: \(4x-3y=7\) => \(4x=3y+7\)
=> \(x=\dfrac{3y+7}{4}\)
=> \(x^2=\left(\dfrac{3y+7}{4}\right)^2\)
=> \(2x^2=\dfrac{\left(3y+7\right)^2}{8}\) (1)
Thay (1) vào B ta có:
B = \(\dfrac{\left(3y+7\right)^2}{8}+5y^2\) = \(\dfrac{9y^2+42y+49+40y^2}{8}\)
= \(\dfrac{49y^2+42y+9+40}{8}\)
= \(\dfrac{\left(7y+3\right)^2}{8}+5\)
Vì \(\dfrac{\left(7y+3\right)^2}{8}\) \(\ge\) 0 => \(\dfrac{\left(7y+3\right)^2}{8}+5\) \(\ge\) 5
=> Dấu bằng xảy ra <=> \(\dfrac{\left(7y+3\right)^2}{8}\) = 0
<=> \(7y+3=0\) <=> \(y=\dfrac{-3}{7}\) => \(x=\dfrac{10}{7}\)
=> GTNN của B = 5 khi \(x=\dfrac{10}{7};y=\dfrac{-3}{7}\)