Question

求帮忙解题

Answer 1

1)
因,x<5/4
所以,4x-5<0
-(4x-5)>0
y=(16x^2-28x+11)/(4x-5)
=[(4x-5)^2+3(4x-5)+1]/(4x-5)
=(4x-5)+[1/(4x-5)]+3
=3-(4x-5)+[1/-(4x-5)]
因-(4x-5)+[1/-(4x-5)] ≥2 √[-(4x-5)*[1/-(4x-5)] =2
当且仅当-(4x-5)=[1/-(4x-5),x=1,时,取等号成立,最小值为:2
所以3-(4x-5)+[1/-(4x-5)] ≥3-2=1

即,y=(16x^2-28x+11)/(4x-5)的最大值为:1
2)
a^2+(1/ab)+[1/a(a-b)]
=[a(a-b)]+ab+(1/ab)+[1/a(a-b)]
={[a(a-b)]+[1/a(a-b)]}+[ab+(1/ab)]≥2√{[a(a-b)]*[1/a(a-b)]}+2√[ab*(1/ab)]=2+2=4
所以最小值为:4