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Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
\(\frac{2}{5}-\frac{1}{7}+\frac{3}{5}.\frac{1}{3}=\frac{14}{35}-\frac{5}{35}+\frac{7}{35}=\frac{16}{35}\)
\(\frac{2}{5}-\frac{1}{7}+\frac{3}{5}\times\frac{1}{3}\)
\(=\frac{2}{5}-\frac{1}{7}+\frac{1}{5}\)
\(=\frac{9}{35}+\frac{1}{5}\)
\(=\frac{16}{35}\)
a) 3/5:6/7-8/15 b)2/3+2/5-1/3 c) 4/9x3/2:2/3 d)1785+564x27
= 7/10 -8/15 =16/15-1/3 =2/3:2/3 =1785+15228
=1/6 =11/15 =1 =17013
a: =11/2*4*5/3
=22*5/3
=110/3
b: =30/12-3/12+20/12
=47/12
c: =28/15+5
=28/15+75/15
=103/15
9900 : 36 - 15 x 11
= 275 - 165
= 110
9700 : 100 + 36 x 12
= 97 + 432
= 529
( 15792 : 336 ) x 5 + 27 x 11
= 47 x 5 + 297
= 235 + 297
= 532
( 160 x 5 - 25 x 4 ) : 4
= ( 800 - 100 ) : 4
= 700 : 4
= 175
1036 + 64 x 52 - 1827
= 1036 + 3328 - 1827
= 4364 - 1827
= 2537
215 . 869 + 215 . 14
= 215 . ( 869 + 14 )
= 215 . 883
= 189945
\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)
= 22 \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{110}{3}\)
\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)
= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5
= \(\dfrac{28}{15}\) + 5
= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)
= \(\dfrac{103}{15}\)
\(=\dfrac{27}{15}-\dfrac{2}{5}=\dfrac{27}{15}-\dfrac{6}{15}=\dfrac{7}{5}\)
=27/15−2/5=27/15−6/15=7/5