先化简 再求值 [（-1/2x^3y^4)^3+(-1/6xy^2)^2*3xy^2]÷(-1/2xy^2)^3 其中x=-2 y=1/2
1个回答
展开全部
[（-1/2x^3y^4)^3+(-1/6xy^2)^2*3xy^2]÷(-1/2xy^2)^3
=[(-1/8)x^9 y^12 + (1/36)x²y^4 * 3xy²] ÷[(-1/8)x³y^6]
=[(-1/8)x^9 y^12 + (1/12)x³y^6] ÷[(-1/8)x³y^6]
=[(-1/8)x^9 y^12]÷[(-1/8)x³y^6] + [(1/12)x³y^6] ÷[(-1/8)x³y^6]
=x^6 y^6 - (2/3)
=(xy)^6 -(2/3)
已知x=-2，y=1/2，那么：xy=-2*(1/2)=-1
所以：原式=(xy)^6 -(2/3)=(-1)^6 - (2/3)=1- (2/3)=1/3
=[(-1/8)x^9 y^12 + (1/36)x²y^4 * 3xy²] ÷[(-1/8)x³y^6]
=[(-1/8)x^9 y^12 + (1/12)x³y^6] ÷[(-1/8)x³y^6]
=[(-1/8)x^9 y^12]÷[(-1/8)x³y^6] + [(1/12)x³y^6] ÷[(-1/8)x³y^6]
=x^6 y^6 - (2/3)
=(xy)^6 -(2/3)
已知x=-2，y=1/2，那么：xy=-2*(1/2)=-1
所以：原式=(xy)^6 -(2/3)=(-1)^6 - (2/3)=1- (2/3)=1/3
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
