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\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)
\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)
\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)
\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-11}{15}\)
\(\dfrac{-9}{8}\)
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
a) \(\dfrac{-4}{3}-\dfrac{2}{7}+\dfrac{4}{3}\)
\(=\left(\dfrac{-4}{3}+\dfrac{4}{3}\right)-\dfrac{2}{7}\)
\(=0-\dfrac{2}{7}=\dfrac{-2}{7}\)
b) \(\dfrac{1}{2}+\dfrac{-4}{5}-\dfrac{3}{10}\)
\(=\dfrac{5}{10}+\dfrac{-8}{10}-\dfrac{3}{10}\)
\(=\dfrac{5+\left(-8\right)-3}{10}=\dfrac{-3}{5}\)
`4 / 15 . 1 / 3 . 15 / 20`
`= [ 4 . 1 . 15 ] / [ 15 . 3 . 20 ]`
`= [ 4 . 1 . 15 ] / [ 15 . 3 . 4 . 5 ]`
`= 1 / [ 3 . 5 ]`
`= 1 / 15`
a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)
\(=\dfrac{3}{5}+\dfrac{41}{25}\)
\(=\dfrac{15}{25}+\dfrac{41}{25}\)
\(=\dfrac{56}{25}\)
a) A = \(\dfrac{3}{5}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) \(\left(\dfrac{45}{4}-\dfrac{37}{4}\right)\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}+\dfrac{41}{6}\) . 2 : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{3}\) : \(\dfrac{25}{3}\)
A = \(\dfrac{3}{5}\) + \(\dfrac{41}{25}\)
A = \(\dfrac{56}{25}\)
a: =9+7=16
b: =11+3/13-2-4/7-5-3/13
=4-4/7
=28/7-4/7=24/7
c: =2/7(5+1/4-3-1/4)=2/7x2=4/7
`(6-2 4/5)*3 1/8-1 3/5:1/4`
`=(6-14/5)*25/8-8/5*4`
`=16/5*25/8-32/5`
`=10-32/5=18/5`
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
1,
a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)
b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)
=\(\dfrac{-2}{15}\)
2,
a, 2x+19=25
=>x=3
b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)
=>x=\(\dfrac{-3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)
\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{-5}{12}\)
b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)
\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)
\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)
\(\dfrac{3}{5}:1,5+20\%-3\dfrac{1}{4}\)
\(=\dfrac{3}{5}:\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{13}{4}\)
\(=\dfrac{2}{5}+\dfrac{1}{5}-\dfrac{13}{4}=\dfrac{3}{5}-\dfrac{13}{4}=\dfrac{12}{20}-\dfrac{65}{20}=\dfrac{-53}{20}\)
\(\dfrac{3}{5}:1,5+20\%-3\dfrac{1}{4}\\ =\dfrac{3}{5}:\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{3}{5}\cdot\dfrac{2}{3}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{2}{5}+\dfrac{1}{5}-\dfrac{13}{4}\\ =\dfrac{3}{5}-\dfrac{13}{4}\\ =\dfrac{12}{20}-\dfrac{65}{20}\\ =\dfrac{-53}{20}\)