Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)

chóng váng

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)

cm nha

đăng ký tranghttps://www.youtube.com/channel/UCdMJRiuo_35tKETQtnAYOBQ để xem thêm nhiều bài tập nha

Dễ k cho mình trước rồi mình làm cho

K phai lop 7 nen k phai lam. Biet dau ma lam 

Mi nói bài toán hay nói tau

Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)

\(=\frac{\sqrt{n+1}}{\sqrt{n}.\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thay n = 1, 2, 3, ..., 2011 vào C ta có:

\(C=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)

Vậy \(C=1-\frac{1}{\sqrt{2012}}.\)

uk xie xie (cảm ơn ) bạn , nhưng mik giải ra lâu r

\(A=\frac{2!+\sqrt{3}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{5}}{4!}+....+\frac{2012!+\sqrt{2013}}{2012!}\)

\(=\frac{2!}{2!}+\frac{\sqrt{3}}{2!}+\frac{3!}{3!}+\frac{\sqrt{4}}{3!}+.....+\frac{2012!}{2012!}+\frac{\sqrt{2013}}{2012!}\)

\(=2012+\left(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+....+\frac{\sqrt{2011}}{2012!}\right)\)

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

haizzzz đệ lm sai rồi kìa =((

đệ sai chỗ nào ah sư phụ?

2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

\(\Rightarrow A< \frac{1}{2}\)

1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)

\(\Rightarrow A< 2\)

Bài 2 tạm thời chưa nghĩ ra :))

\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)

\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(\Rightarrow N< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)

\(N< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)