\(\frac{2!+\sqrt{1}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{9}}{4!}+...+\frac{2012!+\sqrt{2011^2}}{2012!}< 2012\)
chứng minh
\(\frac{2!+\sqrt{1}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{9}}{4!}+...+\frac{2012!+\sqrt{2011^2}}{2012!}< 2012\)
Dễ k cho mình trước rồi mình làm cho
K phai lop 7 nen k phai lam. Biet dau ma lam
Chứng minh
\(\frac{2!+\sqrt{1}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{9}}{4!}+...+\frac{2012!+\sqrt{2011^2}}{2012!}<2012\)
\(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+.....+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}\)
rút gọn giúp mình với
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 : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
CM: \(\frac{2!+\sqrt{1}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{5}}{4!}+...+\frac{2012!+\sqrt{2011^2}}{2012!}\) <2012
\(A=\frac{2!+\sqrt{3}}{2!}+\frac{3!+\sqrt{4}}{3!}+\frac{4!+\sqrt{5}}{4!}+....+\frac{2012!+\sqrt{2013}}{2012!}\)
\(=\frac{2!}{2!}+\frac{\sqrt{3}}{2!}+\frac{3!}{3!}+\frac{\sqrt{4}}{3!}+.....+\frac{2012!}{2012!}+\frac{\sqrt{2013}}{2012!}\)
\(=2012+\left(\frac{\sqrt{3}}{2!}+\frac{\sqrt{4}}{3!}+....+\frac{\sqrt{2011}}{2012!}\right)\)
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
\(S=\sqrt{1+2010^2+\frac{2010^2}{2011^2}}+\frac{2010}{2011}+\sqrt{1+2011^2+\frac{2011^2}{2012^2}}+\frac{2011}{2012}+\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}\)
Tính \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2013\sqrt{2012}+2012\sqrt{2013}}\)
= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)
= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)
= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)
= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)
= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)
CMR :
\(\frac{1}{2}+\frac{1}{3.\sqrt{2}}+\frac{1}{4.\sqrt{3}}+...+\frac{1}{2012.\sqrt{2011}}< 2\)
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow S< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
\(\Rightarrow S< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)
chứng minh A= \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2012\sqrt{2011}}\)<2
chứng minh:
N=\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+......+\frac{1}{2012\sqrt{2011}}< 2\)
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow N< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
\(N< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)