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Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)
\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)
\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)
\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)
\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)
\(=\frac{1.41}{39.3}\)
\(=\frac{41}{117}\)
Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)
\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)
\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0=0\)
c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)
\(=\frac{27}{5}.10=27.2=54\)
\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)
1) \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}\\ =\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}\\ =-\dfrac{5}{4}+\dfrac{6}{5}\\ =-\dfrac{1}{20}\)
2) \(\left(-3,2\right)\cdot\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{1}{2}\)
\(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{7}{2}\\ =\dfrac{3}{4}-\dfrac{22}{15}\cdot\dfrac{2}{7}\\ =\dfrac{3}{4}-\dfrac{44}{105}\\ =\dfrac{139}{420}\)
\(1-\frac{3}{2.10}-\frac{3}{4.15}-\frac{3}{6.20}-\frac{3}{8.25}-...-\frac{3}{198.500}\)
\(=1-\left(\frac{3}{2.10}+\frac{3}{4.15}+\frac{3}{6.20}+...+\frac{3}{198.500}\right)\)
\(=1-\frac{3}{2.5}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=1-\frac{3}{10}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{3}{10}.\left(1-\frac{1}{100}\right)\)
\(=1-\frac{3}{10}.\frac{99}{100}\)
\(=1-\frac{297}{1000}\)
\(=\frac{703}{1000}\)
P/s : Không biết đúng hông nha, làm đại
Bài 1:
a) Ta có: \(\frac{7}{8}\cdot\frac{4}{9}+\frac{1}{14}:\frac{5}{14}\)
\(=\frac{28}{72}+\frac{1}{14}\cdot\frac{14}{5}\)
\(=\frac{28}{72}+\frac{1}{5}\)
\(=\frac{140}{360}+\frac{72}{360}\)
\(=\frac{212}{360}=\frac{53}{90}\)
Bài 2:
a) Ta có: \(\frac{2}{3}x+\frac{1}{4}x=\frac{-22}{27}\)
\(\Leftrightarrow x\cdot\left(\frac{2}{3}+\frac{1}{4}\right)=\frac{-22}{27}\)
\(\Leftrightarrow x\cdot\frac{11}{12}=\frac{-22}{27}\)
\(\Leftrightarrow x=\frac{-22}{27}:\frac{11}{12}=\frac{-22}{27}\cdot\frac{12}{11}=-\frac{8}{9}\)
Vậy: \(x=\frac{-8}{9}\)
a.2/3 + 1/3 - x= 3/5
=> 1-x = 3/5
=> x = 1-3/5= 2/5
b. 1/8/15 - 2/3x = 0,2
=> 2/3x= 23/15 - 1/5= 4/3
=> x= 4/3 : 2/3=2
c. 2/3x -3/2x = 5/12
=> x( 2/3 - 3/2) = 5/12
=> x. -5/6 = 5/12
=> x= 5/12 : -5/6
=> x= -1/2
a) 2/3 + 1/3 - x = 3/5
=> 1 - x = 3/5
=> x = 1 - 3/5
x = 2/5
b) \(1\frac{8}{15}-\frac{2}{3}x=0,2\)
=> 23/15 - 2/3x = 0,2
=> 2/3x = 23/15 - 0,2
2/3x = 1,333333333
=>x = 1,333333333 : 2/3
x = 2
c) ???
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
(x+1/4-1/3).(13/6-1/4)=7/46
(x+1/4-1/3).23/12=7/46
(x+1/4-1/3)=7/46:23/12
(x+1/4-1/3)=7/46.12/23
(x+1/4-1/3)=42/529
x+1/4=42/529+1/3
x+1/4=655/1587
x=655/1587-1/4
x=1033=/6348
vậy x=1033/6348