数学简答题三道

Question

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

解：(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/a+5)
=1/a-1/(a+5)=5/(a+5)a
题目错了,应该是1/a(a+1)+1/(a+1)(a+2)+1/(a+2)(a+3)+1/(a+3)(a+4)+1/(a+4)(a+5)

(3)∵abc=1,
∴a=b=c=1
∴[a/(ab+a+1)]+[b/(bc+b+1)]+(ca+c+1)
=(1/3)+(1/3)+3
=11/3
取特殊值.

Answer 2

....有事一道奇怪的题目，，无能为力

参考资料：砸下嘎

Answer 3

看不明白