tính nhanh:
\(\dfrac{-4}{9}.\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\)
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B=-5/9+8/15+-2/11+4/9+7/15
=>B=(-5/9+4/-9)+(8/15+7/15)-2/11
=>B=(-5/9+-4/9)+1-2/11
=>B=-1+1-2/11
=>B=-2/11
Vậy B=-2/11
Ta có :
B=-5/9+8/15+-2/11+4/9+7/15
=>B=(-5/9+4/-9)+(8/15+7/15)-2/11
=>B=(-5/9+-4/9)+1-2/11
=>B=-1+1-2/11
=>B=-2/11
Vậy B=-2/11
1: \(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}\right)+\dfrac{16}{15}\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)
\(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}+\dfrac{4}{7}-\dfrac{5}{9}\right)=0\)
2: \(=\dfrac{29}{9}\left(15+\dfrac{4}{7}-8-\dfrac{1}{7}+\dfrac{15}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{20}{9}\cdot\left(7\cdot\dfrac{18}{7}\right)=\dfrac{20}{9}\cdot18=40\)
a: \(\dfrac{1}{8}+\dfrac{5}{8}=\dfrac{1+5}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
b: \(\dfrac{1}{15}+\dfrac{4}{15}=\dfrac{1+4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{5}{9}+\dfrac{7}{9}=\dfrac{5+7}{9}=\dfrac{12}{9}=\dfrac{4}{3}\)
d: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
a) 18+58=1+58=68=3418+58=1+58=68=34
b) 115+415=1+415=515=13115+415=1+415=515=13
c) 59+79=5+79=129=4359+79=5+79=129=43
d) 23100+27100=23+27100=50100=12
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
\(\dfrac{-5}{9}\)-\(\left(\dfrac{8}{15}+\dfrac{4}{9}\right)\)+\(\dfrac{7}{15}\)=\(\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\)-\(\left(\dfrac{8}{15}-\dfrac{7}{15}\right)\)=-1-1/15=-16/15.
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
\(\dfrac{-4}{9}.\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\)
\(=\dfrac{-4}{9}.\dfrac{7}{15}+\dfrac{-4}{9}.\dfrac{8}{15}\)
\(=\dfrac{-4}{9}.\left(\dfrac{7}{15}+\dfrac{8}{15}\right)\)
\(=\dfrac{-4}{9}.1\)
\(=\dfrac{-4}{9}\)