计算(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)=?

a^5+5a^4+10a^3+10a^2+5a+1=(a+1)^5
这是杨辉三角型啊
所以
(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)=
(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)+1-1=
(x-1+1)^5-1
=x^5-1

(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)=x^5-1
我们令x-1=t
原式=t^5+5t^4+10t^3+10t^2+5t
=(t^5+5t^4+10t^3+10t^2+5t+1)-1
=(t+1)^5-1
带回原变量得
原式=x^5-1

(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5
(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)=(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1)+1-1=(x-1+1)^5-1=x^5-1

二项式定理：
(x-1)^5+5(x-1)^4+10(x-1)^3+10(x-1)^2+5(x-1）+1=[（x-1)+1]^5=x^5;
所以这题的答案为x^5-1;