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(1- 1/2) . ( 1-1/3) . ( 1-1/4).............. (1-1/1999) . (1- 1/2000)
Câu 1:
a) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\left(1-\frac{1}{1999}\right)\left(1-\frac{1}{2000}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{1998}{1999}\cdot\frac{1999}{2000}=\frac{1}{2000}\)
b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)
=>x+99/100=5/3
=>x=5/3-99/100=203/300
c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
=>5-x=7/3
hay x=8/3
Ngày thứ ba đội đó sửa được số đoạn đường là:
1- (\(\frac{5}{9}+\frac{1}{4}\)) = \(\frac{7}{36}\)(đoạn đường)
Đoạn đường đội đó sửa được số mét là:
7 : 7/36= 36(mét
đáp số : 36 mét
câu 2:
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)+x=\frac{2}{3}\)
\(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)+x=\frac{2}{3}\) \(\left(\frac{11}{12}+\frac{1}{12}-\frac{1}{100}\right)+x=\frac{2}{3}\) \(1-\frac{1}{100}+x=\frac{2}{3}\) \(\frac{99}{100}+x=\frac{2}{3}\) \(x=\frac{2}{3}-\frac{99}{100}\) \(x=\frac{200}{300}-\frac{297}{300}\) \(x=\frac{-97}{300}\)
Bài giải
6 km của đoạn đường cần sửa dài là:
1 - 2/5 - 3/7 = 6/35 ( cả đoạn đường )
Độ dài đoạn đường đội đó đã sửa là:
6 : 6/35 = 35 ( km )
Đáp số : 35 km
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(=\dfrac{3}{5}+\dfrac{41}{25}\)
\(=\dfrac{15}{25}+\dfrac{41}{25}\)
\(=\dfrac{56}{25}\)
a) A = \(\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) \(\left(\dfrac{45}{4}-\dfrac{37}{4}\right)\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) . 2 : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{3}\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{25}\)
A = \(\dfrac{56}{25}\)
a) \(\left(\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.34}+...+\frac{11}{89.100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\left(1-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\frac{99}{100}-x=\frac{2}{3}\)
\(x=\frac{99}{100}-\frac{2}{3}\)
\(x=\frac{97}{300}\)
b) \(\frac{x+1}{9}+\frac{x+3}{7}+\frac{x+5}{5}+\frac{x+7}{3}=a\)
\(\Rightarrow\frac{x+1}{9}+1+\frac{x+3}{7}+1+\frac{x+5}{5}+1+\frac{x+7}{3}+1=a+4\)
\(\frac{x+10}{9}+\frac{x+10}{7}+\frac{x+10}{5}+\frac{x+10}{3}=a+4\)
\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=a+4\)
\(a,\left(\frac{11}{12}+\frac{11}{12\cdot23}+\frac{11}{23\cdot34}+...+\frac{11}{89\cdot100}\right)-x=\frac{2}{3}\)
\(\Rightarrow\left(1-\frac{1}{12}+\frac{1}{12}-\frac{1}{23}+\frac{1}{23}-\frac{1}{34}+...+\frac{1}{89}-\frac{1}{100}\right)-x=\frac{2}{3}\)
\(\Rightarrow1-\frac{1}{100}-x=\frac{2}{3}\)
\(\Rightarrow\frac{99}{100}-x=\frac{2}{3}\)
\(\Rightarrow x=\frac{99}{100}-\frac{2}{3}\)
\(\Rightarrow x=\frac{97}{300}\)
b, k hiểu đề :v
\(\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)
\(=\dfrac{-1}{4.5}+\dfrac{-1}{5.6}+\dfrac{-1}{6.7}+\dfrac{-1}{7.8}+\dfrac{-1}{8.9}+\dfrac{-1}{9.10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{5}+\dfrac{-1}{5}-\dfrac{-1}{6}+\dfrac{-1}{6}-\dfrac{-1}{7}+\dfrac{-1}{7}-\dfrac{-1}{8}+\dfrac{-1}{8}-\dfrac{-1}{9}+\dfrac{-1}{9}-\dfrac{-1}{10}\)
\(=\dfrac{-1}{4}-\dfrac{-1}{10}\)
\(=\dfrac{-3}{20}\)
ban tra loi di