f(x)=(ax^2+1)/(bx+c)(ABC属于Z)为奇函数 f(-x)=(ax^2+1)/(-bx+c)= -f(x)=(ax^2+1)/(-bx-c) -bx+c=-bx-c c=0 f(1)=(a+1)/(b+c)=(a+1)/b=2,a+1=2b f(2)=(4a+1)/(2b+c)=(4a+1)/2b

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f(x)=(ax^2+1)/(bx+c)(ABC属于Z)为奇函数f(-x)=(ax^2+1)/(-bx+c)=-f(x)=(ax^2+1)/(-bx-c)-bx+c=-bx-cc=0f(1)=(a+1

f(x)=(ax^2+1)/(bx+c)(ABC属于Z)为奇函数 f(-x)=(ax^2+1)/(-bx+c)= -f(x)=(ax^2+1)/(-bx-c) -bx+c=-bx-c c=0 f(1)=(a+1)/(b+c)=(a+1)/b=2,a+1=2b f(2)=(4a+1)/(2b+c)=(4a+1)/2b
f(x)=(ax^2+1)/(bx+c)(ABC属于Z)为奇函数
f(-x)=(ax^2+1)/(-bx+c)= -f(x)=(ax^2+1)/(-bx-c)
-bx+c=-bx-c
c=0
f(1)=(a+1)/(b+c)=(a+1)/b=2,a+1=2b
f(2)=(4a+1)/(2b+c)=(4a+1)/2b

f(x)=(ax^2+1)/(bx+c)(ABC属于Z)为奇函数 f(-x)=(ax^2+1)/(-bx+c)= -f(x)=(ax^2+1)/(-bx-c) -bx+c=-bx-c c=0 f(1)=(a+1)/(b+c)=(a+1)/b=2,a+1=2b f(2)=(4a+1)/(2b+c)=(4a+1)/2b
(4a+1)/ 2b < 3
(4a+1)/(a+1)<3
(4a+4-3)/(a+1)<3
[4(a+1)-3]/(a+1)<3
4(a+1)/(a+1)- 3 (a+1)<3
4 - 3(a+1) < 3

什么一二不是