ta có \(x+y\le5=>-\left(x+y\right)\ge-5\)
có \(A=x+y+\dfrac{8}{x}+\dfrac{18}{y}=-\left(x+y\right)+2x+2y+\dfrac{8}{x}+\dfrac{18}{y}\)
có \(-\left(x+y\right)+2x+2y+\dfrac{8}{x}+\dfrac{18}{y}\ge-5+8+12=15\)
=>A\(\ge15\) dấu= xảy ra <=>x=2,y=3
vậy min A=15