2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)

3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)

4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)

6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)

\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)

\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)

\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)

\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)

làm giống như trên

\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)

\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)

P/S: . là nhân nha

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Cái này dễ thế mà ko bít lm

b: \(=8+\dfrac{7}{12}-5-\dfrac{3}{8}=3+\dfrac{5}{24}=\dfrac{77}{24}\)

`a)3/5+(-4/9)`

`=3/5-4/9`

`=27/45-20/45`

`=7/45`

`b)3/5+2/5 . 15/8`

`=3/5 + 30/40`

`=3/5+3/4`

`=12/20+15/20`

`=27/20`

`c)7/2 . 8/13 + 8/13 . (-5/2)`

`=8/13 . (7/2 +(-5)/2)`

`=8/13 . 1`

`=8/13`

`d)-5/17 . (-9/23)+9/23 . (-22/17) + 11 9/23`

`=-5/17 . (-9/23) + 9/23 . (-1) . 22/17 + 11 + 9/23`

`=-5/17 . (-9/23) + (-9/23) . 22/17+11+9/23`

`= -9/23 ( -5/17 + 22/17)+11+9/23`

`= - 9/23 . 1+11+9/23`

`=-9/23+11+9/23`

`=(-9/23+9/23)+11`

`=0+11`

`=11`

a: =27/45-20/45=7/45

b: =3/5+30/40

=3/5+3/4

=12/20+15/20

=27/20

c: =8/13(7/2-5/2+1)=8/13*2=16/13

d: =9/23*5/17-9/23*22/17+11+9/23

=-9/23+11+9/23

=11

56666666

ta có \(\frac{\frac{4}{5}-\frac{4}{11}+\frac{4}{13}}{\frac{9}{5}-\frac{9}{11}+\frac{9}{13}}+\frac{\frac{5}{7}+\frac{5}{11}-\frac{5}{13}}{\frac{9}{7}+\frac{9}{11}-\frac{9}{13}}\)

       \(=\frac{4\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{9\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{5\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}{9\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}\)

           \(=\frac{4}{9}+\frac{5}{9}=\frac{9}{9}=1\)

=7/11+5/6+4/11-1/6

=(7/11+4/11)+(5/6-1/6)

=1+2/3

=5/3

\(A,\dfrac{7}{11}+\dfrac{5}{6}-\dfrac{-4}{11}-\dfrac{1}{6}\)

\(= (\dfrac{7}{11}+\dfrac{-4}{11})-(\dfrac{5}{6}-\dfrac{1}{6})\)

\(=\dfrac{3}{11}-\dfrac{2}{3}\)

\(= \dfrac{9}{33}-\dfrac{22}{33}\)

\(= \dfrac{-13}{33}\)

\(B,\dfrac{5}{9}.\dfrac{7}{3}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}:\dfrac{13}{5}\)

\(=(\dfrac{5}{9}.\dfrac{5}{9}:\dfrac{5}{9}).\dfrac{7}{3}+\dfrac{9}{13}-\dfrac{13}{5}\)

\(=(\dfrac{5}{9}.\dfrac{7}{3})+(\dfrac{9}{13}-\dfrac{13}{5})\)

\(=(\dfrac{5}{9}.\dfrac{21}{9})+(\dfrac{45}{65}-\dfrac{169}{65})\)

\(=\dfrac{35}{27}+\dfrac{-124}{65}\)

\(=\dfrac{2275}{1755}+\dfrac{-3348}{1755}\)

\(=\dfrac{-1073}{1755}\)

a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)

\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)

\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)

b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)

\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)

\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)

c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)

\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)

\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)

d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)

\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)

\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)

\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)

2:

=>2y=30

=>y=15

3:

=1/2(9/7-4/7)+5/7*13/2

=5/7(1/2+13/2)

=5/7*7=5

mik học lớp 5

Ta có : B=(1/5-1/5)-(3/7-3/7)+(5/9-5/9)-(2/11-2/11)+(7/13-7/13)-9/16

B=-9/16