详细过程
设f〔x〕=4∧x÷(4∧x+2)若0<a<1,(1)试求;f(a)+f(1-a)的值。(2)f[1/1001]+f[2/1001]+f[3/1001]+…+f[1000/1001]的值设f〔x〕=4
答案:1 悬赏:50 手机版
解决时间 2021-06-06 19:18
- 提问者网友:眉目添风霜
- 2021-06-06 01:27
最佳答案
- 五星知识达人网友:不如潦草
- 2021-06-06 02:27
解:
(1)f(a)+f(1-a)
=4^a/(4^a+2)+4^(1-a)/[4^(1-a)+2]
=4^a/(4^a+2)+4/[4+2*4^a]
=4^a/(4^a+2)+2/[2+4^a]
=(4^a+2)/[2+4^a]
=1
(2)由(1)可知:
f(1/100)+f(99/100)=1
f(2/100)+f(98/100)=1
...
f(49/100)+f(51/100)=1
f(50/100)+f(50/100)=1
故:
f(1/100)+f(2/100)……+f(99/100)
=[f(1/100)+f(99/100)]+[f(2/100)+f(98/100)]+...+[f(49/100)+f(51/100)]+f(50/100)
=1+1+...+1+4^(1/2)/[4^(1/2)+2]
=49+2/[2+2]
=49.5
6
(1)f(a)+f(1-a)
=4^a/(4^a+2)+4^(1-a)/[4^(1-a)+2]
=4^a/(4^a+2)+4/[4+2*4^a]
=4^a/(4^a+2)+2/[2+4^a]
=(4^a+2)/[2+4^a]
=1
(2)由(1)可知:
f(1/100)+f(99/100)=1
f(2/100)+f(98/100)=1
...
f(49/100)+f(51/100)=1
f(50/100)+f(50/100)=1
故:
f(1/100)+f(2/100)……+f(99/100)
=[f(1/100)+f(99/100)]+[f(2/100)+f(98/100)]+...+[f(49/100)+f(51/100)]+f(50/100)
=1+1+...+1+4^(1/2)/[4^(1/2)+2]
=49+2/[2+2]
=49.5
6
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