\(\frac{2y}{xy}+\frac{3x}{xy}=6\)
=> 2y+3x= 6xy => 3(x+y) = 6xy+y =>3(x+y) = y(6x+1)
=> x+y= \(\frac{y\left(6x+1\right)}{3}\)
Vậy min a= min trên
\(\frac{y\left(6x+1\right)}{3}>=0\) =>min\(\frac{y\left(6x+1\right)}{3}=0\)
=> y(6x+1)=0 => y=0 hoac 6x+1=0 =>6x=-1 => x=-1/6
nhớ thích cho mình nha!!!