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14 tháng 2 2022
\(\dfrac{7}{21}+\dfrac{-9}{36}=\dfrac{1}{3}+\dfrac{-1}{4}=\dfrac{4}{12}+\dfrac{-3}{12}=\dfrac{1}{12}\)
\(\dfrac{-12}{18}+\dfrac{-21}{35}=\dfrac{-2}{3}+\dfrac{-3}{5}=\dfrac{-10}{15}+\dfrac{-9}{15}=\dfrac{-19}{15}\)
\(\dfrac{-18}{14}+\dfrac{15}{-21}=\dfrac{-9}{7}+\dfrac{-5}{7}=\dfrac{-14}{7}=-2\)
\(\dfrac{3}{21}+\dfrac{-6}{42}=\dfrac{1}{7}+\dfrac{-1}{7}=0\)
S
3
27 tháng 2 2022
\(=\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{2}-\dfrac{5}{6}+2022=2022\)
16 tháng 4 2017
Tính các tổng dưới đây sau khi đã rút gọn phân số :
a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
8 tháng 1
a: \(\dfrac{5}{7}=\dfrac{5\cdot11}{7\cdot11}=\dfrac{55}{77}\)
\(\dfrac{9}{11}=\dfrac{9\cdot7}{11\cdot7}=\dfrac{63}{77}\)
b: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot9}{7\cdot9}=\dfrac{54}{63}\)
\(-\dfrac{12}{54}=\dfrac{-2}{9}=\dfrac{-2\cdot7}{9\cdot7}=-\dfrac{14}{63}\)
c: \(\dfrac{-11}{30}=\dfrac{-11\cdot4}{30\cdot4}=\dfrac{-44}{120}\)
\(\dfrac{-17}{-40}=\dfrac{17}{40}=\dfrac{17\cdot3}{40\cdot3}=\dfrac{51}{120}\)
d: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot3}{7\cdot3}=\dfrac{18}{21}\)
\(\dfrac{-12}{36}=\dfrac{-1}{3}=\dfrac{-1\cdot7}{3\cdot7}=\dfrac{-7}{21}\)
PD
0
18 tháng 7 2021
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
18 tháng 7 2021
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
29 tháng 3
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
7 tháng 8
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
\(-\dfrac{18}{42}\cdot\dfrac{35}{42}:\dfrac{-98}{42}\)
\(=-\dfrac{18}{42}\cdot\dfrac{35}{42}\cdot\dfrac{42}{-98}\)
\(=\dfrac{-18}{42}\cdot\dfrac{-5}{14}\\ =\dfrac{15}{98}\)
\(\dfrac{-18}{42}.\dfrac{3\text{5}}{42}:\dfrac{-98}{42}\)
\(=\dfrac{-18}{42}.\dfrac{3\text{5}}{42}.\dfrac{-42}{98}\)
\(=\dfrac{1\text{5}}{98}\)