1 phần 1×3 + 1phần 2×4 + 1phần 3×5+.......+1phần +98×100
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MV
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\\ =\left(2-1\right)\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ =1-\dfrac{1}{2^{99}}< 1\)
Vậy \(B< 1\)
17 tháng 7 2017
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\)
\(\Rightarrow2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\dfrac{1}{2^{99}}\)
\(\rightarrow B< 1\rightarrowđpcm\)
16 tháng 5 2021
7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0
7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)
7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)
7/48 - 193/56 : x = 0
193/56 : x = 0 + 7/48
193/56 : x = 7/48
x = 193/56 : 7/48
x = 1158/49
NH
2
8 tháng 5 2021
\(\frac{3}{2}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\): \(\frac{1}{5}\)= \(\frac{9}{8}\)x \(\frac{4}{5}\): \(\frac{1}{5}\)= \(\frac{9}{10}\): \(\frac{1}{5}\)= \(\frac{9}{10}\)x \(\frac{5}{1}\)= \(\frac{9}{2}\)
TP
2
K
6 tháng 5 2022
\(\dfrac{2}{3}\times\left(x+\dfrac{4}{5}\right)=\dfrac{-1}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}:\dfrac{2}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}\times\dfrac{3}{2}\\ x+\dfrac{4}{5}=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}-\dfrac{4}{5}\\ x=\dfrac{-5}{10}-\dfrac{8}{10}\\ x=\dfrac{-13}{10}\)
6 tháng 5 2022
\(\dfrac{2}{3}.\left(x+\dfrac{4}{5}\right)=-\dfrac{1}{3}\)
\(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}:\dfrac{2}{3}\right)\)
\(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}.\dfrac{3}{2}\right)=-\dfrac{1}{2}\)
\(x\) \(=\left(-\dfrac{1}{2}\right)-\dfrac{4}{5}\)
\(x\) \(=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{4}{5}\right)\)
\(x\) \(=-\dfrac{13}{10}\)
LT
0
LP
2 tháng 1 2022
\(\dfrac{2x+1}{3}=\dfrac{4}{5}\Leftrightarrow5\left(2x+1\right)=4.3\Leftrightarrow10x+5=12\Leftrightarrow x=\dfrac{7}{10}\)
10 tháng 2 2022
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{20}{27}\)
hay x=20/27+1/4=107/108
10 tháng 2 2022
\(\dfrac{x-1}{4}=\dfrac{5}{3}.\dfrac{4}{9}\Leftrightarrow\dfrac{x-1}{4}=\dfrac{20}{27}\Rightarrow27x-27=80\)
\(\Leftrightarrow27x=107\Leftrightarrow x=\dfrac{107}{27}\)
Bài làm:
Ta có: \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{98}{99}+\frac{1}{2}.\frac{49}{100}\)
\(=\frac{49}{99}+\frac{49}{200}\)
\(=\frac{14651}{19800}\)