Question

f(2x 1)=4x²-6x-5

Answer 1

换元法令2x+1=t,得2x=t-1,代入
f(t)=(t-1)^2-3(t-1)-5=t^2-5t-1,
即f(x)=x^2-5x-1

Answer 2

解：设2X+1=t，则
ⅹ=(t-1)/2，
∴f(t)=4[(t-1)/2]²-6(t-1)/2-5
=t²-2t+1-3t+3-5
=t²-5t-1，
∴f(X)=X²-5X-1

Answer 3

f(2x+1)
=4x^2-6x-5
=2(2x+1)^2 -14x -7
=2(2x+1)^2 -7(2x+1)
ie
f(x) =2x^2 -7x

Answer 4

方法一，换元法:
设2X+1=t，则
ⅹ=(t-1)/2，
∴f(t)=4[(t-1)/2]²-6(t-1)/2-5
=t²-2t+1-3t+3-5
=t²-5t-1，
∴f(X)=X²-5X-1。
方法二，配凑法:
f(2X+1)=4x²-6ⅹ-5
=(2Ⅹ+1)²-10X-6
=(2X+1)²-5(2x+1)-1，
∴f(X)=X²-5X-1。