1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

1) Tính biểu thức:

a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)

b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)

=9/17+8/17=1

c: =>x-3/10=7/15*1/5=7/75

=>x=7/75+3/10=59/150

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

Giải chi tiết cho mình nha

\(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=\dfrac{22}{31}\)

\(\dfrac{16}{24}:4=\dfrac{16}{24}\times\dfrac{1}{4}=\dfrac{4}{24}=\dfrac{1}{6}\)

\(30:\dfrac{6}{5}=30\times\dfrac{5}{6}=25\)

\(\dfrac{9}{16}:4=\dfrac{9}{16}\times\dfrac{1}{4}=\dfrac{9}{64}\)

\(\dfrac{1}{2}:\dfrac{1}{3}\times\dfrac{8}{15}=\dfrac{1}{2}\times3\times\dfrac{8}{15}=\dfrac{4}{5}\)

2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5

Đề bài : Tìm x???

\(\left(8-x+5-9\right)=\left|3-7+9-15\right|-\left(8-4+5\right)\)

\(8-x+5-9=\left|-10\right|-9\)

\(8-x+5-9=1\)

\(8-x+5=10\)

\(8-x=5\)

\(x=3\)

Vậy \(x=3\)

a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)

\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)

=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)

b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)

Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)

c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)

\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)

=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)

d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)

\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6

=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)

a) 3/5x7/8:10/9=21/40:10/9=189/400

b) 3 2/7x2 1/4:1 2/5=23/7x9/4:7/5=207/28:7/5=1035/196

c) 6/9+24/15+12/9+6/15=34/15+12/9+6/15=18/5+6/15=4

d) 3 2/9-( 1 5/8+1 2/9 )=29/9-(13/8+11/9)=29/9-205/72=3/8

tinh

a)  3/5 x 7/8 : 10/9

b) 3 2/7 x 2 1/4 : 1 2/5

c) 6/9 + 24/15 + 12/9 + 6/15

a,x=9/10

b,x=16/3

c,1/7

d,1205/741

e,313/175

bạn ơi bạn bị sai mất 2 câu cuối rồi đó làm ơn tính lại giùm mình nha

a: \(\dfrac{-8}{3}\cdot\dfrac{15}{24}=\dfrac{-8}{24}\cdot\dfrac{15}{3}=\dfrac{-1}{3}\cdot5=-\dfrac{5}{3}\)

b: \(=\dfrac{3}{4}\cdot\dfrac{1}{-9}=\dfrac{-1}{12}\)

c: \(x=-\dfrac{7}{6}+\dfrac{5}{8}=-\dfrac{13}{24}\)

d: \(x=-\dfrac{14}{25}-\dfrac{3}{4}=\dfrac{-56-75}{100}=\dfrac{-131}{100}\)