Question

一道高一数学题，求高手解答

Answer 1

f(x)=1/√(x^2-4)
所以x=1/√(y^2-4)
x^2=1/(y^2-4)```````y＜2 且x＞0
因为在该点上
(an)^2=1/(1/(an-1)^2-4)
(an)^2=1/(1/(an-1)^2-4)
1/(an)^2=1/(an-1)^2-4
所以是等差数列`````
第二问
a1=1
设cn=1/(an)^2
cn=(bn-1)-4
c1=1
所以cn=4n-3
1/(an)^2=4n-3
x＞0
所以an=1/√(4n-3)
第三小问
bn=1/(√(4n-3)+√(4n+1))
分母有理化
bn=(√(4n+1)-√ (4n-3))/4
所以sn=(1/4)(√(4n+1)-√(4n-3)+√(4n-3)-√(4n-7)```````+√5-√1)
sn=(1/4)(√(4n+1)-1)
打了很多字````望加分`````++++望采纳`````````望悬赏`````字符好难打啊```还有口算`````

Answer 2

0.0