Question

急急急！一道二次根式的题

若a=25，求下式的值： (1/√a+√a+1)+(1/√a+1+√a+2)+.....+(1/√a+599+√a+600) 也就是说：1除以根号a加根号下a加1，加上1除以根号下a加1加上根号下a加2.....
你们都化简怎么都没算答案。a=25

Answer 1

有理化1/(√a+√a+1)=[√a-√(a+1)]/[√a+√a+1][√a-√a+1]=√a+1-√a

(1/√a+√a+1)+(1/√a+1+√a+2)+.....+(1/√a+599+√a+600)
=(√a+1-√a)+(√a+2-√a+1)+...+(√a+600-√a+599)
=√(a+600)-√a

Answer 2

(1/√a+√a+1)+(1/√a+1+√a+2)+.....+(1/√a+599+√a+600)
=√(a+1)－√a＋√(a+2)－√(a+1)+.....+√(a+600)－√(a+599)
=√(a+600)-√a