Thực hiện phép tính:\(\frac{-2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\frac{2018}{2019}\)
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Ta có : \(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)=\frac{2018}{2017}-2019.2-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(2018.\left(\frac{1}{2017}-\frac{2019}{1009}\right)-2019.\left(\frac{1}{2017}-2\right)\)
\(=\frac{2018}{2017}-2018.\frac{2019}{1009}-\frac{2019}{2017}+2019.2\)
\(=\frac{2018}{2017}-2.2019-\frac{2019}{2017}+2.2019\)
\(=\frac{2018}{2017}-\frac{2019}{2017}=-\frac{1}{2017}\)
\(a,=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}\)
\(=\frac{11}{16}:\frac{11}{32}\)
\(=\frac{11}{16}.\frac{32}{11}\)
\(=2\)
Đặt \(2018=a\)
\(\Rightarrow\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(=\sqrt{\frac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}=\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2019\)
Chúc mày học ngu
Chúc mày học ngu
Chúc mày học ngu
Chúc mày học ngu
\(\left(\frac{19}{2018}-2019\right).\frac{1}{2019}-\left(\frac{1}{2018}-2019\right).\frac{19}{2019}\)
\(=\frac{19}{2018}-2019.\frac{1}{2019}-\frac{-1}{2018}+2019.\frac{19}{2019}\)
\(=\left(\frac{19}{2018}-\frac{-1}{2018}\right)-\left(2019+2019\right).\left(\frac{1}{2019}.\frac{19}{2019}\right)\)
\(=\frac{18}{2018}-2038.\frac{19}{2019}\)
còn đâu tự tính nha
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
\(-\frac{2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\frac{2018}{2019}=\frac{2018}{2019}\left(\frac{-2-5}{7}\right)+1\frac{2018}{2019}=\frac{2018}{2019}.\left(-1\right)+1\frac{2018}{2019}=\frac{-2018}{2019}+1\frac{2018}{2019}=1\)
Cảm ơn bạn