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a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
\(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{1}{7}.\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{1}{7}.\dfrac{-15}{2}+\dfrac{5}{2}.\dfrac{3}{7}\)
=\(\dfrac{-15}{14}+\dfrac{15}{14}\)
= 0
a ) 4 - \(1\dfrac{4}{5}\).\(\dfrac{-3}{4}\) = 4 - \(\dfrac{-3}{5}\)= \(\dfrac{23}{5}\)
b) \(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}:\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\)
\(\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}:\dfrac{3}{7}-\dfrac{59}{6}\cdot\dfrac{1}{7}\)
\(\dfrac{1}{7}\cdot\left(\dfrac{7}{3}-\dfrac{59}{6}\right)+\dfrac{35}{6}\)
\(\dfrac{1}{7}\cdot\dfrac{-15}{2}+\dfrac{35}{6}=\dfrac{-15}{14}+\dfrac{35}{6}=\dfrac{100}{21}\)
\(a,4-1\dfrac{5}{7}.\left(-0,75\right)\) = \(4-\dfrac{12}{7}.\dfrac{-75}{100}\) = \(4-\dfrac{12}{7}.\dfrac{3}{4}\)
=\(4-\dfrac{36}{28}=4-\dfrac{9}{7}=\dfrac{28}{7}-\dfrac{9}{7}=\dfrac{19}{7}\)
b, \(\dfrac{1}{7}.2\dfrac{1}{3}+\dfrac{5}{2}:\dfrac{3}{7}-\dfrac{59}{6}.\dfrac{1}{7}\) \(=\dfrac{1}{7}.\dfrac{7}{3}+\dfrac{5}{2}.\dfrac{7}{3}-\dfrac{59}{6}.\dfrac{1}{7}=\dfrac{7}{3}.\left(\dfrac{1}{7}+\dfrac{5}{2}\right)-\dfrac{59}{6}.\dfrac{1}{7}\)
=\(\dfrac{7}{3}.\left(\dfrac{2}{14}+\dfrac{35}{14}\right)-\dfrac{59}{42}=\dfrac{7}{3}.\dfrac{37}{14}-\dfrac{59}{42}\)
= \(\dfrac{1.37}{3.2}-\dfrac{59}{42}=\dfrac{37}{6}-\dfrac{59}{42}=\dfrac{259}{42}-\dfrac{59}{42}=\dfrac{200}{42}=\dfrac{100}{21}\)
\(a.\dfrac{2}{3}+\dfrac{4}{3}:\dfrac{-2}{3}=\dfrac{2}{3}+\left(-2\right)=\dfrac{-4}{3}\)
\(b.3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\\ =3\dfrac{4}{5}-2\dfrac{1}{4}-1\dfrac{4}{5}\\ =\left(3\dfrac{4}{5}-1\dfrac{4}{5}\right)-2\dfrac{1}{4}\\ =2-2\dfrac{1}{4}=\dfrac{1}{4}\)
\(c.\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\\ =\dfrac{-3}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
\(=\dfrac{2}{5}+\dfrac{4}{3}.\dfrac{-3}{2}\)
\(=\dfrac{2}{5}+-2\)
\(=\dfrac{2}{5}+\dfrac{-10}{5}\)
\(=\dfrac{-8}{5}\)
-6/8=-3/2
2/7
Quy đồng được 12/20+-35/20=-23/20
Quy đồng được -10/15+3/15=-7/15
Quy đồng lên được -4/26+-5/26=-9/26
Quy đồng lên được: -12/21+7/21=-5/21 nhé
\(\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-6}{8}=\dfrac{-3}{4}\)
\(\dfrac{-3}{7}+\dfrac{5}{7}=\dfrac{2}{7}\)
\(\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{12}{20}+\dfrac{-35}{20}=\dfrac{-23}{20}\)
\(\dfrac{-2}{3}+\dfrac{1}{5}=\dfrac{-10}{15}+\dfrac{3}{15}=\dfrac{-13}{15}\)
\(\dfrac{2}{13}+\dfrac{-5}{26}=\dfrac{-4}{26}+\dfrac{-5}{26}=\dfrac{-9}{26}\)
\(\dfrac{-4}{7}+\dfrac{1}{3}=\dfrac{-12}{21}+\dfrac{7}{21}=\dfrac{-5}{21}\)
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}+\dfrac{\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}}=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}\right)}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}\)
\(=1\)
\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)(đã sửa)
Đặt \(A=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\right)\)
\(\dfrac{2}{3}A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}:\dfrac{2}{3}=\dfrac{56}{305}\cdot\dfrac{3}{2}=\dfrac{84}{305}\)
ST trả lời đúng rồi đó!!!