f(x)=1/(2^x+根号2),求f(-8)+f(-7)+…+f(0)+…+f(8)+f(9) 用倒序相加法求我想要较为详细的步骤,就是很通俗易懂的,每一个步骤,
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![f(x)=1/(2^x+根号2),求f(-8)+f(-7)+…+f(0)+…+f(8)+f(9) 用倒序相加法求我想要较为详细的步骤,就是很通俗易懂的,每一个步骤,](/uploads/image/z/14735835-27-5.jpg?t=f%28x%29%3D1%2F%EF%BC%882%5Ex%2B%E6%A0%B9%E5%8F%B72%EF%BC%89%2C%E6%B1%82f%28-8%29%2Bf%28-7%29%2B%E2%80%A6%2Bf%280%29%2B%E2%80%A6%2Bf%288%29%2Bf%289%29+%E7%94%A8%E5%80%92%E5%BA%8F%E7%9B%B8%E5%8A%A0%E6%B3%95%E6%B1%82%E6%88%91%E6%83%B3%E8%A6%81%E8%BE%83%E4%B8%BA%E8%AF%A6%E7%BB%86%E7%9A%84%E6%AD%A5%E9%AA%A4%2C%E5%B0%B1%E6%98%AF%E5%BE%88%E9%80%9A%E4%BF%97%E6%98%93%E6%87%82%E7%9A%84%2C%E6%AF%8F%E4%B8%80%E4%B8%AA%E6%AD%A5%E9%AA%A4%2C)
f(x)=1/(2^x+根号2),求f(-8)+f(-7)+…+f(0)+…+f(8)+f(9) 用倒序相加法求我想要较为详细的步骤,就是很通俗易懂的,每一个步骤,
f(x)=1/(2^x+根号2),求f(-8)+f(-7)+…+f(0)+…+f(8)+f(9) 用倒序相加法求
我想要较为详细的步骤,
就是很通俗易懂的,每一个步骤,
f(x)=1/(2^x+根号2),求f(-8)+f(-7)+…+f(0)+…+f(8)+f(9) 用倒序相加法求我想要较为详细的步骤,就是很通俗易懂的,每一个步骤,
f(t)+f(1-t)=1/(2^t+根号2)+1/(2^(1-t)+根号2) 后面的分式分子分母同乘以 2^t
=1/(2^t+根号2)+2^t/(2^+根号2* 2^t)
=根号2/(根号2*2^t+2)+2^t/(2^+根号2* 2^t)
=1
S=f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)
S=f(9)+f(8)+.+f(1)+.+f(-7)+f(-8)
相加,对应的和都是1
2S=1*18
S=9
f(x)=1/(2^x+√2)
f(-x)=1/[2^(-x)+√2]=1/[1/2^(x)+√2]=1/{[1+√2*2^(x)]/2^(x)}=2^(x)/[1+√2*2^(x)]
=√2*2^(x)/[√2+2*2^(x)]=√2*2^(x)/[√2+2^(x+1)]
所以f(x+1)+f(-x)=1/[2^(x+1)+√2]+√2*2^(x)/[√2+2^(x+1)...
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f(x)=1/(2^x+√2)
f(-x)=1/[2^(-x)+√2]=1/[1/2^(x)+√2]=1/{[1+√2*2^(x)]/2^(x)}=2^(x)/[1+√2*2^(x)]
=√2*2^(x)/[√2+2*2^(x)]=√2*2^(x)/[√2+2^(x+1)]
所以f(x+1)+f(-x)=1/[2^(x+1)+√2]+√2*2^(x)/[√2+2^(x+1)]=[1+√2*2^(x)]/[2^(x+1)+√2]=√2/2
S=f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)
S=f(9)+f(8)+.......+f(1)+....+f(-7)+f(-8)
2S=√2/2*18
S=9√2/2
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