（x²-x+y）^5的展开式中x³y²项的系数等于

Question

如图，希望有过程，谢谢

Answer 1

(x^2-x+y)^5
=(x^2-x)^5 +5(x^2-x)^4.y + 10(x^2-x)^3.y^2
+10(x^2-x)^2.y^3 +5(x^2-x).y^4 + y^5
coef. x^3.y^2 of "(x^2-x+y)^5"
=coef. x^3.y^2 of "10(x^2-x)^3.y^2"
=coef. x^3.y^2 of "10x^3. (x-1)^3.y^2"
=-10本回答被提问者采纳