\(\Rightarrow\)C=\(\frac{\text{1.2.3.4...99.100}}{\text{2.4.6....100}}\) 

\(\Rightarrow\)C=\(\frac{\text{1.2.3...99.100}}{(2.2....2)(1.2.3.4.5....50)}\) [50 chữ số 2]
\(\Rightarrow\)C=\(\frac{51}{2}.\frac{52}{2}...\frac{100}{2}\)=D

vậy C=D

Mk thấy khó hiểu , bn có thể chỉ kĩ hơn đc ko.

 \(\frac{1}{2^2}>\frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(....\)

\(\frac{1}{2015^2}>\frac{1}{2014.2015}=\frac{1}{2014}-\frac{1}{2015}\)

nên \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2015^2}>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2004}-\frac{1}{2005}\)

\(=1-\frac{1}{2005}\)

vì \(1-\frac{1}{2005}< 1\)

=> ĐPCM

NHân 3 rồi tìm S đi

Xem câu hỏi tương tự í

Thành viên OLM dạo này lười thế nhờ có cái đó cũng phải hỏi 

• đáp án :S=1

\(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:\frac{1}{4}\)

\(=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\)

\(=\frac{16}{5}.\frac{25}{8}-\frac{32}{5}\)

\(=10-\frac{32}{5}\)

\(=\frac{18}{5}\)

bạn có chép sai đề ko

a)\(\left(x+\frac{1^2}{4}\right)=\frac{4}{9}\)

\(x+\frac{1}{16}=\frac{4}{9}\)

\(x=\frac{4}{9}-\frac{1}{16}\)

\(x=\frac{55}{144}\)

b)\(\left(2x-1^2\right)=16\)

\(2x-1=16\)

\(2x=16+1\)

\(2x=17\)

\(x=17:2=\frac{17}{2}\)

\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

\(\frac{3}{x}=\frac{-1}{2}\)

\(\Rightarrow3.2=\left(-1\right).x\)

\(\Rightarrow6=\left(-1\right).x\)

\(\Rightarrow x=6:\left(-1\right)\)

\(\Rightarrow x=-6\)

\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)

\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)

\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)

\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)

\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)

Bài 1:

a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) ta có: \(A=1+2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2019}\)

\(\Rightarrow2A-A=2^{2019}-2\)

\(\Rightarrow A=2^{2019}-2\)

Chúc bn học tốt !!!!!

a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Bạn nên kiểm tra kĩ lại đề.

Đúng đề mà bn, ko sai đc đâu, mk chắc chắn mà.