Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=(-1/4.5)+(-1/5.6)+(-1/6.7)+(-1/7.8)+(-1/8x9)+(-1/9.10)
A=(-1/4)-(-1/5)+(-1/5)-(-1/6)+(-1/6)-(-1/7)+(-1/7)-(-1/8)+(-1/8)-(-1/9)-(-1/9)+(-1/10)
A=(-1/4)-(-1/10)
A=-1/4+1/10
A=-3/20
`[-1]/2+[-1]/6+[-1]/12+[-1]/20+[-1]/30+[-1]/42+[-1]/56+[-1]/72+[-1]/90`
`=(-1)(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=(-1)(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)`
`=(-1)(1-1/10)`
`=(-1). 9/10=-9/10`
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{6}\)+ \(\dfrac{-1}{12}\)+ \(\dfrac{-1}{20}\)+ \(\dfrac{-1}{30}\)+ \(\dfrac{-1}{42}\)+ \(\dfrac{-1}{56}\)+ \(\dfrac{-1}{72}\)+ \(\dfrac{-1}{90}\)
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{2\times3}\)+ \(\dfrac{-1}{3\times4}\)+ \(\dfrac{-1}{4\times5}\)+ \(\dfrac{-1}{5\times6}\)+ \(\dfrac{-1}{6\times7}\)+ \(\dfrac{-1}{7\times8}\)+ \(\dfrac{-1}{8\times9}\)+ + \(\dfrac{-1}{9\times10}\)
A = - (\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+ \(\dfrac{1}{9}\)-\(\dfrac{1}{10}\))
A = -(1-\(\dfrac{1}{10}\))
A = \(\dfrac{-9}{10}\)
=9/10-(1/2+1/6+...+1/90)
=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)
=9/10-9/10=0
Quãng sông AB dài là :
8 giờ 24 phú x 10 = 84 (km)
Vận tốc cua ca nô khi xuôi dòng là :
10 + 2 = 12 (km/giờ )
Thời gian ca nô đi xuôi dòng là :
84 : 2 = 7 (giờ )
Đáp số : 7 giờ
dễ mà
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}=\dfrac{9}{10}\)
\(2\dfrac{2}{9}-x=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{9}\)
\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{2}{9}\)
\(\Rightarrow x=2\dfrac{2}{9}-\dfrac{2}{9}\)
\(\Rightarrow x=2\)
Vậy x=2
2\(\dfrac{2}{9}\) - x = \(\dfrac{1}{3\cdot4}\)+\(\dfrac{1}{4\cdot5}\)+\(\dfrac{1}{5\cdot6}\)+\(\dfrac{1}{6\cdot7}\)+\(\dfrac{1}{7\cdot8}\)+\(\dfrac{1}{8\cdot9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{9}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{9}{27}\)- \(\dfrac{3}{27}\)
2\(\dfrac{2}{9}\)-x = \(\dfrac{2}{9}\)
\(\dfrac{20}{9}\) -x = \(\dfrac{2}{9}\)
x = \(\dfrac{20}{9}-\dfrac{2}{9}\)
x = 2
Vậy x = 2
Ta có:
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(x=\dfrac{41}{40}\)
\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)
\(\Leftrightarrow\dfrac{7}{30}-x=\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{7}{30}-\dfrac{-19}{24}\)
\(\Rightarrow x=\dfrac{41}{40}\)
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
Link này bạn:Câu hỏi của Hoàng Hà Nhi - Toán lớp 6 | Học trực tuyến
\(\dfrac{x-1}{12}+\dfrac{x-1}{20}+\dfrac{x-1}{30}+...+\dfrac{x-1}{72}=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\dfrac{2}{9}=\dfrac{16}{9}\\ x-1=8\\ x=8+1\\ x=9\)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\\ =\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\\ =\dfrac{1}{3}-\dfrac{1}{9}\\ =\dfrac{2}{9}\)
\(a,\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(=\dfrac{1}{212}\)