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a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
a: \(=\dfrac{15}{4}:\dfrac{2-7}{16}-\dfrac{5}{9}\cdot\left(\dfrac{3}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{15}{4}\cdot\dfrac{16}{-5}-\dfrac{5}{9}\cdot\dfrac{7}{5}\)
\(=\dfrac{-240}{20}-\dfrac{7}{9}=-12-\dfrac{7}{9}=\dfrac{-115}{9}\)
c: \(=\dfrac{1}{5}-\dfrac{7}{2}-\dfrac{3}{2}+\dfrac{5}{4}\)
\(=\dfrac{4+25}{20}-5=\dfrac{29}{20}-\dfrac{100}{20}=\dfrac{-71}{20}\)
d: \(=\dfrac{12}{17}\left(1-\dfrac{1}{15}-\dfrac{4}{5}+1\right)=\dfrac{12}{17}\cdot\dfrac{17}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
e: \(=\dfrac{2}{15}\cdot\dfrac{9}{8}-\dfrac{7}{4}\cdot\dfrac{9}{8}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\left(\dfrac{2}{15}-\dfrac{7}{4}\right)+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{8-105}{60}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{-97}{60}+\dfrac{25}{36}=\dfrac{-1619}{1440}\)
a: \(\dfrac{7}{2}-1,5+\dfrac{-8}{11}=3,5-1,5-\dfrac{8}{11}=2-\dfrac{8}{11}=\dfrac{22-8}{11}=\dfrac{14}{11}\)
b: \(\left(1\dfrac{3}{7}+3\dfrac{4}{7}\right):\dfrac{6}{5}-2,5\cdot\dfrac{8}{15}\)
\(=\left(1+\dfrac{3}{7}+3+\dfrac{4}{7}\right)\cdot\dfrac{5}{6}-\dfrac{5}{2}\cdot\dfrac{8}{15}\)
\(=5\cdot\dfrac{5}{6}-\dfrac{40}{30}=\dfrac{25}{6}-\dfrac{4}{3}=\dfrac{17}{6}\)
c: \(\dfrac{7}{9}\cdot\dfrac{35}{17}-\dfrac{7}{9}\cdot\dfrac{25}{17}-\dfrac{7}{9}\)
\(=\dfrac{7}{9}\left(\dfrac{35}{17}-\dfrac{25}{17}-1\right)\)
\(=\dfrac{7}{9}\cdot\dfrac{-7}{17}=\dfrac{-49}{117}\)
d: \(\left(120\%+1\dfrac{3}{5}\right):\dfrac{27}{5}-4,5\cdot\dfrac{4}{9}\)
\(=\left(1,2+1,6\right)\cdot\dfrac{5}{27}-2\)
\(=\dfrac{14}{27}-2=\dfrac{14}{27}-\dfrac{54}{27}=-\dfrac{40}{27}\)
e: \(\dfrac{-5}{7}\cdot x=\dfrac{25}{12}\)
=>\(x=-\dfrac{25}{12}:\dfrac{5}{7}=-\dfrac{25}{12}\cdot\dfrac{7}{5}=\dfrac{-5\cdot7}{12}=-\dfrac{35}{12}\)
g: \(\dfrac{7}{11}+\dfrac{1}{11}\cdot x=0,3-3\dfrac{2}{5}\)
=>\(\dfrac{x}{11}=0,3-3,4-\dfrac{7}{11}=-\dfrac{411}{110}\)
=>\(x=-\dfrac{411}{110}\cdot11=-\dfrac{411}{10}\)
hơi dài đó.