设f(2x+1)=5x+3,则f(x)=( )a 5x+1 b (5/2)x+1 c (5/2)x+1/2 d 5x+1/2

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设f(2x+1)=5x+3,则f(x)=( )a 5x+1 b (5/2)x+1 c (5/2)x+1/2 d 5x+1/2
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设f(2x+1)=5x+3,则f(x)=( )a 5x+1 b (5/2)x+1 c (5/2)x+1/2 d 5x+1/2
设f(2x+1)=5x+3,则f(x)=( )
a 5x+1 b (5/2)x+1 c (5/2)x+1/2 d 5x+1/2

设f(2x+1)=5x+3,则f(x)=( )a 5x+1 b (5/2)x+1 c (5/2)x+1/2 d 5x+1/2
把2x+1看成一个整体,5x是2x的2.5倍,所以2.5*(2x+1)=5x+2.5.(5x+3)-(5x+2.5)=0.5.设(2x+1)=t.f(t)=2.5t+0.5.所以f(x)=2.5x+0.5.答案是c

设2x+1=t,则x=(t-1)/2 f(2x+1)=f(t)=5(t-1)/2+3=5/2t+1/2 再将x 回代即得f(x)=5/2x+1/2 所以答案为c (5/2)x+1/2