2x+3y=5
=>x=\(\frac{5-3x}{2}\)
=>F=\(2.\frac{\left(5-3y\right)^2}{4}+3y^2=\frac{25-30y+9y^2}{2}+\frac{6y^2}{2}\)
\(=\frac{25-30y+15y^2}{2}=\frac{15y^2-30y+15+10}{2}\)
\(=\frac{15.\left(y-1\right)^2+10}{2}=\frac{15.\left(y-1\right)^2}{2}+5\ge5\)
Dấu "=" xảy ra khi : y=1 =>x=\(\frac{5-3}{2}=1\)
kakaka bik giải rùi