4.5642548e+74

 Số 0 sai rồi 12^x không bao giờ có số tận cùng là 0 đâu

1,0008969x10^16

k nha

12 x 23 x 34 x 45 x 56 x 67 x 78 x 89 x 910 = 10008969

a/ x = 118

b/ x = 134

c/ 54x - 44 = 12457986

=> 54x = 12458030

=> x = 230704,2593

Smile thèm l i k e

\(\frac{87}{89}< \frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}< \frac{88}{45}\)

Đặt \(A=\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}\)

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

\(\Rightarrow1-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(\Rightarrow1-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2011}\)

\(\Rightarrow\frac{88}{45}>\frac{2011-\sqrt{2011}}{2011}>A>\frac{2010}{2011}>\frac{87}{89}\)

\(\Rightarrow\frac{87}{89}< \frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}< \frac{88}{45}\)

Phạm Thị Diệu Huyền chà chà gắt quá

tổng quát nhé \(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}=\frac{1}{\sqrt{\left(k+1\right)k}}>\frac{1}{\left(k+1\right)\sqrt{k}}>\frac{1}{\left(k+1\right)k}=\frac{1}{k}-\frac{1}{k+1}\)
Đặt A= \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+...+\frac{1}{2011\sqrt{2010}}\)
\(\Rightarrow1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)\(\Leftrightarrow1-\frac{1}{\sqrt{2011}}>A>1-\frac{1}{2011}\Leftrightarrow\frac{88}{45}>\frac{2011-\sqrt{2011}}{2011}>A>\frac{2010}{2011}>\frac{87}{89}\)

ban chung minh tong quat ro hon ko

Xét với n là số tự nhiên không nhỏ hơn 1 , ta có 

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

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

Áp dụng điều trên : 

\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}< \)

\(< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2010}}\right)=2\left(1-\frac{1}{\sqrt{2010}}\right)< \)

\(< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)