biểu thức đâu

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

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

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

\(C =( 49\dfrac{8}{23} - 4\dfrac{8}{23}) - 5\dfrac{7}{32}\)

\(C = 45 - 5\dfrac{7}{32}\)

\(C = \dfrac{1273}{32}\)

Trời ơi cái đề bài !!!

Thoy thì làm từng câu vậy

a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)

\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\dfrac{7}{222222}\)

\(I=\dfrac{7}{22}\)

1.

a) \(A=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)\(A=\dfrac{10101.5}{10101.11}+\dfrac{10101.5}{10101.22}-\dfrac{10101.4}{10101.11}\)

\(A=\dfrac{5}{11}+\dfrac{5}{22}-\dfrac{4}{11}=\dfrac{7}{22}\)

\(B=\dfrac{2698}{45}-\left(\dfrac{344}{45}-\dfrac{17}{57}\right)\)

(Áp dụng tính chất a - (b - c) = a-b-c)

\(B=\dfrac{2698}{45}-\dfrac{344}{45}-\dfrac{17}{57}\)

\(B=\dfrac{2354}{45}-\dfrac{17}{57}\)

\(B=\dfrac{44726}{855}-\dfrac{255}{855}\)

\(B=\dfrac{44471}{855}\)

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

\(=30-\dfrac{89}{736}=\dfrac{21991}{736}\)

=30−89/736=21991/736

= 21991/736

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

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

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

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

\(=\dfrac{1135}{23}-\left(\left(5+14\right)+\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\right)\)

\(=\dfrac{1135}{23}-\left(19+\dfrac{417}{736}\right)\)

\(=\dfrac{1135}{23}-19\dfrac{417}{736}\)

\(=\dfrac{1135}{23}-\dfrac{14401}{736}\)

\(=\dfrac{953}{32}\)

b) \(-\dfrac{3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=-\dfrac{1}{7}\cdot\dfrac{5}{3}-\dfrac{4}{3}\cdot\dfrac{1}{7}+\dfrac{17}{7}\)

\(=-\dfrac{5}{21}-\dfrac{4}{21}+\dfrac{17}{7}\)

\(=2\)

c) \(0,7\cdot2\dfrac{2}{3}\cdot20\cdot0,375\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=2\cdot\dfrac{5}{4}\)

\(=\dfrac{5}{2}\)

d) \(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(=\left(9\cdot\dfrac{3}{8}+\dfrac{303030}{69264}\right)+\dfrac{403}{100}\)

\(=\left(\dfrac{27}{8}+\dfrac{35}{8}\right)+\dfrac{403}{100}\)

\(=\dfrac{31}{4}+\dfrac{403}{100}\)

\(=\dfrac{589}{50}\)

P/s: Đánh dấu phẩy, dấu chấm (dấu nhân) cần rõ ràng (vì dấu chấm người ta sẽ hiểu là dấu nhân thay vì hiểu là dấu phẩy)

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

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

=35+\(5\dfrac{7}{32}\)

=\(\dfrac{1287}{32}\)

b)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).\dfrac{9}{9}\right]+2\dfrac{3}{7}\)

=\(\left[\left(\dfrac{-3}{7}\right).1\right]+2\dfrac{3}{7}\)

=\(\left(\dfrac{-3}{7}\right)+2\dfrac{3}{7}\)

=2

c) 0,7.\(2\dfrac{2}{3}\).20.0.375.\(\dfrac{5}{28}\)

=0 (Vì có một thừ số là 0 nên nguyên cả tích là 0)

d)\(\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

=17+4,03

=21,03

(mk ko ghi lại đề đâu nhé)

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

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

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

= 34 +\(\dfrac{32}{32}-\dfrac{7}{32}\)

= 34 + \(\dfrac{25}{32}\) = 34\(\dfrac{25}{32}\)

B = (6+5) + (\(\dfrac{3}{8}+\dfrac{1}{2}\))

= 11 + \(\dfrac{7}{8}\)

= 11\(\dfrac{7}{8}\)

C = \(8\dfrac{2}{7}-3\dfrac{4}{9}-4\dfrac{2}{7}\)

= \(8\dfrac{2}{7}-4\dfrac{2}{7}-3\dfrac{4}{9}\)

= 4 - 3\(\dfrac{4}{9}\)

= \(3+\dfrac{9}{9}-3-\dfrac{4}{9}\)

= \(\dfrac{5}{9}\)