H
hocngoan291
Bài 1 câu 2:
A= 1/([tex]\sqrt{1}+\sqrt{3})+ 1/(\sqrt{5}+\sqrt{7})+...+1/(\sqrt{97}+\sqrt{99}[/tex])>9/4
Vì 1/([tex]\sqrt{n}+\sqrt{n+2})>1/(\sqrt{n}+\sqrt{n+4}[/tex])
=> 1/([tex]\sqrt{1}+\sqrt{3}[/tex])> 1/([tex]\sqrt{1}+\sqrt{5}[/tex]) = [tex]\sqrt{5}- \sqrt{1}[/tex]tương tự như vậy ta có: 1/([tex]\sqrt{5}+\sqrt{7}[/tex])> 1/([tex]\sqrt{5}+\sqrt{9}[/tex]) = [tex]\sqrt{9}- \sqrt{5}[/tex]
=>A>[tex] \sqrt{5}- \sqrt{1}+\sqrt{9}- \sqrt{5}+ \sqrt{13}- \sqrt{9}+...+ \sqrt{101}- \sqrt{97}[/tex]
=>A> [tex]\sqrt{101}- \sqrt{1}[/tex]
=>A> 9/4
A= 1/([tex]\sqrt{1}+\sqrt{3})+ 1/(\sqrt{5}+\sqrt{7})+...+1/(\sqrt{97}+\sqrt{99}[/tex])>9/4
Vì 1/([tex]\sqrt{n}+\sqrt{n+2})>1/(\sqrt{n}+\sqrt{n+4}[/tex])
=> 1/([tex]\sqrt{1}+\sqrt{3}[/tex])> 1/([tex]\sqrt{1}+\sqrt{5}[/tex]) = [tex]\sqrt{5}- \sqrt{1}[/tex]tương tự như vậy ta có: 1/([tex]\sqrt{5}+\sqrt{7}[/tex])> 1/([tex]\sqrt{5}+\sqrt{9}[/tex]) = [tex]\sqrt{9}- \sqrt{5}[/tex]
=>A>[tex] \sqrt{5}- \sqrt{1}+\sqrt{9}- \sqrt{5}+ \sqrt{13}- \sqrt{9}+...+ \sqrt{101}- \sqrt{97}[/tex]
=>A> [tex]\sqrt{101}- \sqrt{1}[/tex]
=>A> 9/4
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