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ăn lồn cái địt mẹ mày lửa trùa ĐẦU LỒN nhá

pro ĐẦU LỒN mày nhá

a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)

nên -11<x<-8

hay \(x\in\left\{-10;-9\right\}\)

b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)

Suy ra: 9<x<14

hay \(x\in\left\{10;11;12;13\right\}\)

c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)

nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)

Suy ra: -536<x<-63

hay \(x\in\left\{-535;-534;...;-64\right\}\)

Bài làm :

\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)

\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)

\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)

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: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

`# \text {DNamNgV}`

`16/21 + 6/7`

`= 16/21 + 18/21`

`= 34/21`

__

`15/22 \times 11/35`

`= (15 \times 11)/(22 \times 35)`

`= (5 \times 3 \times 11)/(2 \times 11 \times 5 \times 7)`

`= (3 \times 1)/(2 \times 7)`

`= 3/14`

___

`8/11 \div 5/22`

`= 8/11 \times 22/5`

`= (8 \times 22)/(11 \times 5)`

`= (8 \times 11 \times 2)/(11 \times 5)`

`= (8 \times 2)/5`

`= 16/5`

___

`9/13 \div 27/39`

`= 9/13 \times 39/27`

`= 9/13 \times 13/9`

`= 1`

8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)

9: =2/7-2/5+5/7=1-2/5=3/5

10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)

11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)

8 = -2/11

9 = 3/5

10 = -5/19

11 = -5/7

11 = 5/13

a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)

=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)

=\(\left(-1\right)+1+\dfrac{-2}{11}\)

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

b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)

=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)

=\(0+0+\dfrac{7}{17}\)

=\(\dfrac{7}{17}\)

c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)

A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)

A=\(35-5\dfrac{7}{32}\)

A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)

d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)

C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)

a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`

`=-5/9+8/15-2/11-4/9+7/15`

`=(-5/9-4/9)+(8/15+7/15)-2/11`

`=-9/9+15/15-2/11`

`=-1+1-2/11`

`=-2/11`

b, `((-5)/12+6/11)+(7/17+5/11+5/12)`

`=-5/12+6/11+7/17+5/11+5/12`

`=(-5/12+5/12)+(6/11+5/11)+7/17`

`=0+11/11+7/17`

`=1+7/17`

`=17/17+7/17`

`=24/17`

c, `A=49 8/23 - (5 7/32 + 14 8/23)`

`A=49 8/23 - 5 7/32 - 14 8/23`

`A=(49 8/23 - 14 8/23)-5 7/32`

`A=35 - 167/32`

`A=953/32`

d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`

`C=-3/7 . 5/9-4/9 . 3/7+17/7`

`C=-3/7.(5/9+4/9)+17/7`

`C=-3/7 . 1+17/7`

`C=2`

a) Ta có: \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{-4}{9}+\dfrac{7}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}\)

\(=-1+1-\dfrac{2}{11}\)

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

b) Ta có: \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{11}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)

\(=\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{7}{11}+\dfrac{5}{11}\right)\)

\(=\dfrac{18}{11}\)

c) Ta có: \(A=49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)

\(=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}\)

\(=30-\dfrac{7}{32}=\dfrac{953}{32}\)