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\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)
\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)
\(=\frac{19}{4}+\left(-\frac{83}{20}\right)+\frac{37}{12}=\frac{3}{5}+\frac{37}{12}=\frac{221}{60}\)
\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)
\(\frac{19}{4}-\frac{37}{100}+\frac{1}{8}-\frac{32}{25}-\frac{5}{2}+\frac{7}{2}=\left(\frac{19}{4}-\frac{37}{100}-\frac{32}{25}\right)+\left(\frac{7}{2}-\frac{5}{2}\right)+\frac{1}{8}=\frac{31}{10}+1+\frac{1}{8}=\frac{169}{40}\)
(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)
=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10
=8/15.48/63.120/143.3/8.35/48.9/10
=384/945.360/1144.315/480
=138240/1081080.315/480
=43545600/518918400=84/1001
a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
Ta có :
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{10^2}\right)\)
\(=\)\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.....\frac{10^2-1}{10^2}\)
\(=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{9.11}{10.10}\)
\(=\)\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)
\(=\)\(\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}\)
\(=\)\(\frac{11}{2.10}\)
\(=\)\(\frac{11}{20}\)
Chúc bạn học tốt ~
(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/10^2)
=3/4.8/9.15/16...99/100
Từ đó tính kết quả là ok
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
\(=\frac{19}{4}+\left(-\frac{37}{100}\right)+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{219}{50}+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{25}{10}\right)+\frac{35}{10}\)
\(=\frac{41}{10}\)
\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{2}\)
\(=\frac{19}{4}-\frac{37}{100}-\frac{128}{100}-\frac{25}{10}+\frac{7}{2}\)
\(=\frac{475}{100}-\frac{37}{100}-\frac{128}{100}-\frac{250}{100}+\frac{350}{100}\)
\(=\frac{410}{100}=4,1\)