已知f（x+1）=2x方-3x+1 求f（x）为啥到了最后一步能把t换成x呢 那里我不懂

设 x+1 = t, x=t-1
f（x+1）=2x^2-3x+1
f(t)=2(t-1)^2-3(t-1)+1
=2(t^2-2t+1)-3t+3+1
=2t^2 -4t+2-3t+4
=2t^2-7t+6
所以 f(x)=2x^2-7x+6

令x+1=t 则x=t-1
把t-1代入式子中的x有:
f(t)=2(t-1)²-3(t-1)+1
=2(t²-2t+1)-3t+4
=2t²-7t+6
f(t)等价于f(x) 即t等价于x
所以f(x)=2x²-7x+6

设X=x+1,得x=X-1.带入式中得：
f(X)=2(X-1)^2-3(X-1)+1
=2(X^2-2X+1)-3X+3+1
=2X^2-4X+2-3X+4
=2X^2-7X+6
即f(x)=2x^2-7x+6

f(x+1)=2x^2-3x+1=2[(x+1)-2]^2+(x+1)-2
f(x)=2(x-2)^2+x-2=2x^2-7x+6