Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{1+2+3+4+5+6+7+8+9}{10}\)
\(=\frac{45}{10}=\frac{9}{2}\)
a,
\(=\frac{1+2+3+4+5+6+7+8+9}{10}=\frac{45}{10}=\frac{9}{2}\)
b,\(=9\times\left(\frac{26\times10101}{27\times10101}+\frac{8\times11111}{9\times11111}\right)=9\times\left(\frac{26}{27}+\frac{8}{9}\right)=9\times\frac{50}{27}=\frac{50}{3}\)
k cho mk nha!!!!!!!!!!!!!
Mk cần gấp các bạn ơi !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!?
a) $\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$
b) $\frac{2}{3} - \frac{4}{{15}} = \frac{{10}}{{15}} - \frac{4}{{15}} = \frac{6}{{15}} = \frac{2}{5}$
c) $\frac{3}{5} - \frac{{10}}{{25}} = \frac{{15}}{{25}} - \frac{{10}}{{25}} = \frac{5}{{25}} = \frac{1}{5}$
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)