Làm giúp mk với:
X=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+\(\frac{1}{1+2+3+4}\)+........+\(\frac{1}{1+2+3+4+....+20}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{200}}{\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+....+\left(\frac{198}{2}+1\right)+1}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}}{\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}}=\frac{1}{2000}\)
\(\left(x-20\right)\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)}=\frac{1}{2000}\)
\(\left(x-20\right).\frac{1}{200}=\frac{1}{2000}\)
\(\left(x-20\right)=\frac{1}{2000}:\frac{1}{200}=\frac{1}{2000}.200=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+20=\frac{201}{10}\)
a, Đề có vẻ sai sai nhé :v
b, \(\left|\frac{1}{2}x-\frac{2}{3}\right|-1=\frac{1}{6}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{2}{3}\right|=\frac{7}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{2}{3}=\frac{7}{6}\\\frac{1}{2}x-\frac{2}{3}=-\frac{7}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{3}\\x=-1\end{cases}}\)
Vậy : ....
c, \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{4}{5}\)
\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\cdot(x+1)}=\frac{4}{5}\)
\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{4}{5}\)
\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)
\(\Leftrightarrow x+1=10\Leftrightarrow x=9\)
Vậy x = 9
Đề: X=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+.......+\(\frac{1}{1+2+3+4+20}\)
X=\(\frac{1}{2.3:2}\)+\(\frac{1}{3.4:2}\)+\(\frac{1}{4.5:2}\)+......+\(\frac{1}{20.21:2}\)
X=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)\(\frac{2}{4.5}\)+........+\(\frac{2}{20.21}\)
X=2.(\(\frac{1}{2}\).3+\(\frac{1}{3}\).4+\(\frac{1}{4}\).5+.....+\(\frac{1}{20}\).21)
X=2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{20}\)-\(\frac{1}{21}\))
X=2.(\(\frac{1}{2}\)-\(\frac{1}{21}\))
X=2.(\(\frac{21}{42}\)-\(\frac{2}{42}\))
X=2.\(\frac{19}{42}\)
X=\(\frac{19}{21}\)
Mn xem thử đúng ko nha!
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .......... ; \(1+2+3+....+20=\frac{20.21}{2}\)
\(\Rightarrow X=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.......+\frac{1}{\frac{20.21}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+........+\frac{2}{20.21}=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{20.21}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{20}-\frac{1}{21}\right)=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)