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a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha
.
Theo đề : Ta có : 1 .2 + 2.3 + 3.1 + a . 6 + 5.3 + 6.1 +7.4 + 8.2 + 9.1 +10 .2 / 25 = 5,16
=> 1 .2 + 2.3 + 3.1 + a . 6 + 5.3 + 6.1 +7.4 + 8.2 + 9.1 +10 .2 = 5,16 . 25
=> 1 .2 + 2.3 + 3.1 + a . 6 + 5.3 + 6.1 +7.4 + 8.2 + 9.1 +10 .2 = 129
=> a.6 + (1 .2 + 2.3 + 3.1 + 5.3 + 6.1 +7.4 + 8.2 + 9.1 +10 .2) = 129
=> a .6 + 105 = 129
=> a .6 = 129 - 105
=> a .6 = 24
=> a = 24 : 6
=> a = 4
Vậy a = 4
a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)
\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)
\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)
b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)
\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)
\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)
c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)
\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)
\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)
d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4