Question

已知函数f(x)=1/(2x+√2),求f(-5)+f(-4)+~~~+f(0)+~~f(5)+f(6)的值

Answer 1

已知函数 f(x)=1/(2x+√2) ，求f(-5)+f(-4)+~~~+f(0)+~~f(5)+f(6)的值
f(x) = 1/(2x+√2)
f( - x) = 1/(- 2x+√2)
f(x) + f( - x) = 1/(2x+√2) + 1/(- 2x+√2) = 2√2 / (2 - 4x²) ①
f(-5)+f(-4)+~~~+f(0)+~~f(5)+f(6)
= [f(±1) + f(±2) + f(±3) + f(±4) + f(±5)] + f(0) + f(6)
= 2√2 [1/(2-4*1²) + 1/(2-4*2²) + 1/(2-4*3²) + 1/(2-4*4²) + 1/(2-4*5²)] + 1/√2 + 1/(12+√2)
= - 2√2 [1/2 + 1/14 + 1/34 + 1/62 + 1/98] + 1/√2 + 1/(12+√2)
= - 2√2 * 577/920 + √2/2 + (12-√2)/142
= 6 / 71 - 986√2/1295
≈ - 0.99226

Answer 2

(x) = 1/(2x+√2)
f( - x) = 1/(- 2x+√2)
f(x) + f( - x) = 1/(2x+√2) + 1/(- 2x+√2) = 2√2 / (2 - 4x²) ①
f(-5)+f(-4)+~~~+f(0)+~~f(5)+f(6)
= [f(±1) + f(±2) + f(±3) + f(±4) + f(±5)] + f(0) + f(6)
= 2√2 [1/(2-4*1²) + 1/(2-4*2²) + 1/(2-4*3²) + 1/(2-4*4²) + 1/(2-4*5²)] + 1/√2 + 1/(12+√2)
= - 2√2 [1/2 + 1/14 + 1/34 + 1/62 + 1/98] + 1/√2 + 1/(12+√2)
= - 2√2 * 577/920 + √2/2 + (12-√2)/142
= 6 / 71 - 986√2/1295
≈ - 0.99226

Answer 3

f(x)可化为 f(x)=(2x-√2)/(4x^2-2)
所以f(x)+f(-x)=-2√2/(4x^2-2)
把x=5 ,4,3,2,1,0代入 再加个f(6)即可

虽然多 但 不难算 因为x没有根号了