数学简答题三道如果x/y=2,求代数式x^2+y^2/xy.计算:1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+1/(a+4)(a+5)+1/(a+4)(a+5)已知abc=1,求(a/ab+a+1)+(b/bc+b+1)+(ca+c+1)

Question

数学简答题三道
如果x/y=2,求代数式x^2+y^2/xy.
计算:1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+1/(a+4)(a+5)+1/(a+4)(a+5)
已知abc=1,求(a/ab+a+1)+(b/bc+b+1)+(ca+c+1)

Answer 1

(1)∵x/y=2,∴x=2y∴(x^2+y^2)/xy=(4y^2+y^2)/(4y^2)=5y^2/4y^2=5/4(2):1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)]+1/[(a+4)(a+3)]+1/[(a+4)(a+5)]=(1/a-1/a+1)+(1/a+1-1/a+2)+(1/a+2-1/a+3)+(1/a+3-1/a+4)+(1/a+4-1...