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a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
a) MC :24
\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{1\times8+3\times3-7\times2}{24}=\frac{3}{24}=\frac{1}{8}\)
b)MC : 56
\(\frac{3}{14}+\frac{5}{8}-\frac{1}{2}=\frac{3\times4+5\times7-1\times28}{56}=\frac{19}{56}\)
c) MC: 36
\(\frac{1}{4}-\frac{2}{3}-\frac{11}{18}=\frac{1\times9-2\times12-11\times2}{36}=\frac{-37}{36}\)
d) MC: 312
\(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}=\frac{1\times78+5\times26-1\times24-7\times39}{312}=\frac{-89}{312}\)
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
\(=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\)
\(=\frac{10}{30}+\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\)
\(=\frac{20}{30}=\frac{2}{3}\)
tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Áp dụng công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\) nhé bạn
Ta có :
\(A=\frac{2}{1+2}+\frac{2}{1+2+3}+\frac{2}{1+2+3+4}+...+\frac{2}{1+2+3+...+2018}\)
\(A=\frac{2}{\frac{2\left(2+1\right)}{2}}+\frac{2}{\frac{3\left(3+1\right)}{2}}+\frac{2}{\frac{4\left(4+1\right)}{2}}+...+\frac{2}{\frac{2018\left(2018+1\right)}{2}}\)
\(A=\frac{4}{2.3}+\frac{4}{3.4}+\frac{4}{4.5}+...+\frac{4}{2018.2019}\)
\(A=4\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(A=4\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=4\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(A=4.\frac{2017}{4038}\)
\(A=\frac{4034}{2019}\)
Vậy \(A=\frac{4034}{2019}\)
Chúc bạn học tốt ~