F.2 Maths Questions @ fksnlix的部落格

標題:

F.2 Maths Questions

發問:

1) [(1/64)(x^2)(y^(n-1))]+[(1/9)(y^n+1)]-[(1/12)(x)(y^n)] =? 2) (a^2-ab)^2-(ab-b^2)^2-2[a^4-b^4-(a-b)(a+b)^3] =? Thx a lot!!!

最佳解答:

1)[(1/64)(x^2)(y^(n-1))]+[(1/9)(y^n+1)]-[(1/12)(x)(y^n)] =(y^n-1)[(1/64)(x^2)+(1/9)(y^2)-(1/12)(xy)] =(579)(y^n-1)[9(x^2)+64(y^2)-48(xy)] =(579)(y^n-1)[9(x^2)-48(xy)+64(y^2)] =(579)(y^n-1)[(3y-4x)^2] 2)(a^2-ab)^2-(ab-b^2)^2-2[a^4-b^4-(a-b)(a+b)^3] =(a^2-ab-ab+b^2)(a^2-ab+ab-b^2)-2[a^4-b^4-(a-b)(a+b)^3] =(a^2-2ab+b^2)(a^2-b^2)-2[a^4-b^4-(a-b)(a+b)^3] =(a^2-2ab+b^2)(a^2-b^2)-2[(a^2+b^2)(a^2-b^2)-(a-b)(a+b)^3] =(a^2-2ab+b^2)(a^2-b^2)-2[(a^2+b^2)(a-b)(a+b)-(a-b)(a+b)^3] =(a^2-2ab+b^2)(a^2-b^2)-2(a-b)(a+b)[(a^2+b^2)-(a+b)^2] =(a^2-2ab+b^2)(a^2-b^2)-2(a-b)(a+b)[(a^2+b^2)-a^2-2ab-b^2] =(a^2-2ab+b^2)(a^2-b^2)-2(a^2-b^2)[(a^2+b^2)-a^2-2ab-b^2] =(a^2-2ab+b^2)(a^2-b^2)-2(a^2-b^2)[-2ab] =(a^2-2ab+b^2)(a^2-b^2)+4(a^2-b^2)[ab] =(a^2+b^2)(a^2-b^2)-2ab(a^2-b^2)+4ab(a^2-b^2) =(a^2+b^2)(a^2-b^2)+2ab(a^2-b^2) =(a^2+2ab+b^2)(a^2-b^2) =(a+b)^2(a^2-b^2) 好難好難!!!!!!!!