设f(x)=1/(2x+根号2),求f(-5)+f(-4)+…+f(0)……+f(5)+f(6)=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 03:28:10
![设f(x)=1/(2x+根号2),求f(-5)+f(-4)+…+f(0)……+f(5)+f(6)=?](/uploads/image/z/660446-62-6.jpg?t=%E8%AE%BEf%28x%29%3D1%2F%282x%2B%E6%A0%B9%E5%8F%B72%29%2C%E6%B1%82f%28-5%29%2Bf%28-4%29%2B%E2%80%A6%2Bf%280%29%E2%80%A6%E2%80%A6%2Bf%285%29%2Bf%286%29%3D%3F)
xJ@_,n УXcQ*AR<5al
dvo7_&~~>.Znk"ի+\=fTfzL2}9blRR6YII7c*Yp2 \_VOp>-
设f(x)=1/(2x+根号2),求f(-5)+f(-4)+…+f(0)……+f(5)+f(6)=?
设f(x)=1/(2x+根号2),求f(-5)+f(-4)+…+f(0)……+f(5)+f(6)=?
设f(x)=1/(2x+根号2),求f(-5)+f(-4)+…+f(0)……+f(5)+f(6)=?
f(x)+f(-x)=-根号2/(2x^2-1) 将x=12345分别带入后想加得到了f(5)+f(-5)+f(4)+f(-4)...+f(1)+f(-1),f(0)=1/根号2,f(6)=1/(12+根号2),最后结果就出来了,1/根号2 +1/(12+根号2)-根号2(1+1/7+1/17+1/31+1/49)
f(1)+f(-1)=1/(2+根号2)+1/(根号2-2)=-根号2,
f(2)+f(-2)=-根号2/7
f(3)+f(-3)=-根号2/17
......
f(-5)+f(-4)+…+f(0)……+f(5)+f(6)
=[f(-5)+f(5)]+[f(-4)+f(-4)]+…+[f(-1)+f(1)]+f(0)+f(6)