求f(x)= 1/x + 1/(2-x) (0
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求f(x)= 1/x + 1/(2-x) (0
求f(x)= 1/x + 1/(2-x) (0
求f(x)= 1/x + 1/(2-x) (0
00.
∴1/x+1/(2-x)
=1²/x+1²/(2-x)
≥(1+1)²/[x+(2-x)]
=2,
故所求最小值为:2.
此时,x=1.
用高一所学的基本不等式解,则
1/x+1/(2-x)
=[1/x+1/(2-x)]·1
=[1/x+1/(2-x)]·[x+(2-x)]/2
=(2-x)/x+x/(2-x)+2
≥2√[x/(2-x)·(x-2)/x]+2
=4.
故所求最小值为:4.
此时,由(2-x)/x=x/(x-2)得x=1.
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