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ta có:
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{2018}}\)
\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2018}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2017}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+........+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2018}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2018}}\)
\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)
a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)
Đến đây bn tự tính nhé !!
\(\left(\frac{2}{3}-1\frac{1}{2}\right):\frac{4}{3}+\frac{1}{2}\)
=\(\left(\frac{2}{3}-\frac{3}{2}\right)\times\frac{3}{4}+\frac{1}{2}\)
=\(\frac{-5}{6}\times\frac{3}{4}+\frac{1}{2}\)
=\(\frac{-5}{8}+\frac{4}{8}\)
=\(\frac{-1}{8}\)
Ai thấy đúng thì *******
\(\left(\frac{2}{3}-1\frac{1}{2}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\left(\frac{2}{3}-\frac{3}{2}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\left(\frac{4}{6}-\frac{9}{6}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\frac{-5}{6}:\frac{4}{3}+\frac{1}{2}\)
\(=\frac{-5}{6}.\frac{3}{4}+\frac{1}{2}\)
\(=\frac{-5}{8}+\frac{1}{2}\)
\(=\frac{-5}{8}+\frac{4}{8}\)
\(=\frac{1}{8}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2018}{2017}.\frac{2019}{2018}\)
\(=\frac{3.4.5.6.....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2019}{2018}\)
\(=\frac{3.4.5....2019}{2.3.4....2018}\)
\(=\frac{2019}{2}\)