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-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](/uploads/image/z/6080393-65-3.jpg?t=f%28x%29%3D%28ax%5E2%2B1%29%2F%28bx%2Bc%29%28ABC%E5%B1%9E%E4%BA%8EZ%29%E4%B8%BA%E5%A5%87%E5%87%BD%E6%95%B0+f%28-x%29%3D%28ax%5E2%2B1%29%2F%28-bx%2Bc%29%3D+-f%28x%29%3D%28ax%5E2%2B1%29%2F%28-bx-c%29+-bx%2Bc%3D-bx-c+c%3D0+f%281%29%3D%28a%2B1%29%2F%28b%2Bc%29%3D%28a%2B1%29%2Fb%3D2%2Ca%2B1%3D2b+f%282%29%3D%284a%2B1%29%2F%282b%2Bc%29%3D%284a%2B1%29%2F2b)
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