Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{2!+\sqrt{3}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{5}}{4!}+....+\frac{2012!+\sqrt{2013}}{2012!}\)
\(=\frac{2!}{2!}+\frac{\sqrt{3}}{2!}+\frac{3!}{3!}+\frac{\sqrt{4}}{3!}+.....+\frac{2012!}{2012!}+\frac{\sqrt{2013}}{2012!}\)
\(=2012+\left(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+....+\frac{\sqrt{2011}}{2012!}\right)\)
Mà \(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+...+\frac{\sqrt{2013}}{2012!}>0\)
\(\Rightarrow A>2012+0=2012\)
Đề sai nên t sửa lại r nhé
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
______________________________________________
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
Giải:
(1+1/2!)+(1+2/3!)+(1+3/4!)+....+(1+2011/2012!)=2011+(1/2!+2/3!+3/4!+...+2011/2012!)
=2011+(\(\frac{1}{2!}\)+\(\frac{3-1}{3!}\)+\(\frac{4-1}{4!}\)+...+\(\frac{2012-1}{2012!}\))= 2011 +(\(\frac{1}{2!}\)+\(\frac{1}{2!}\)-\(\frac{1}{3!}\)+\(\frac{1}{3!}\)-\(\frac{1}{4!}\)+...+\(\frac{1}{2011!}\)-\(\frac{1}{2012!}\))
= 2011+(1-\(\frac{1}{2012!}\))=2012 - \(\frac{1}{2012!}\)<2012 (đpcm)
Đề đánh sai rồi kìa =.=
Ở dưới ít hơn 1 căn mà.