高中数学不等式,三道题,详细步骤

2025-03-20 04:01:29
推荐回答(1个)
回答1:

4:把y看成自变量
x=1/y,代入:
f(y)=(1/y²+4y²-4+4)/(1/y-2y)
=((1/y-2y)²+4)/(1/y-2y)
=(1/y-2y)+4/(1/y-2y)
0√2<1/y<+∞,
0<2y<√2,-√2<-2y<0
0<1/y-2y<+∞
f(y)=(1/y-2y)+4/(1/y-2y)-4+4
=【√(1/y-2y)-2/√(1/y-2y)】²+4
≥4
满足下面的条件时等号成立:

√(1/y-2y)=2/√(1/y-2y)
1/y-2y=2
1-2y²=2y
2y²+2y-1=0
y²+y=1/2
y²+y+1/4=3/4
(y+1/2)²=3/4
y+1/2=√3/2
y=(√3-1)/2