\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

Câu a :  

\(\dfrac{5}{12}+\dfrac{3}{4}+\dfrac{1}{3}\\ =\dfrac{5}{12}+\dfrac{9}{12}+\dfrac{4}{12}=\dfrac{3}{2}\)

câu b : 

\(\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{11}{14}\\ =\dfrac{7}{28}+\dfrac{12}{28}+\dfrac{22}{28}=\dfrac{41}{28}\)

câu c : 

\(\dfrac{1}{4}+\dfrac{3}{12}+\dfrac{12}{36}\\ =\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{4}{12}\\ =\dfrac{10}{12}=\dfrac{5}{6}\)

địt con mẹ mày lên

\(\dfrac{1}{3}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)=\(\dfrac{3}{15}\)-\(\dfrac{2}{15}\)+\(\dfrac{14}{15}\)

=\(\dfrac{13}{15}\)

\(\dfrac{3}{5}\)+4-\(\dfrac{6}{7}\)=\(\dfrac{3}{5}\)+\(\dfrac{20}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{23}{5}\)-\(\dfrac{6}{7}\)=\(\dfrac{161}{35}\)-\(\dfrac{30}{35}\)=\(\dfrac{131}{35}\)

Bài 1:

\(\left(\frac{9}{4}:\frac{6}{5}\right)-1=\frac{45}{24}-1=\frac{15}{8}-1=\frac{7}{8}\)

\(2+\left(\frac{1}{4}\times\frac{2}{5}\right)=2+\frac{2}{10}=2+\frac{1}{5}=\frac{11}{5}\)

Bài 2: 

\(\frac{3}{5}+\frac{2}{3}=\frac{9}{15}+\frac{10}{15}=\frac{19}{15}\)

\(1-\frac{9}{11}=\frac{1}{1}-\frac{9}{11}=\frac{11}{11}-\frac{9}{11}=\frac{2}{11}\)

\(\frac{7}{9}\times\frac{3}{14}=\frac{21}{126}=\frac{3}{18}\)

\(\frac{15}{7}:\frac{5}{21}=\frac{315}{35}=9\)

1 Tính giá trị biểu thức

( 9/4 : 6/5 ) - 1 = 45/24 - 1 = 7/8

2 + ( 1/4 × 2/5 ) = 2 + 1/10 = 21/10

2 Tính

3/5 + 2/3 = 

1 - 9/11 = 2/11

7/9 × 3/14 = 1/6

15/7 : 5/21 = 9

(9/4 : 6/5) - 1 = (9/4 . 5/6) - 1

= 15/8 - 1 = 7/8

2 + (1/4 x 2/5)  = 2 + 1/10

= 21/10

3/5 + 2/3 = 19/15

1 - 9/11 = 2/11

7/9 x 3/14 = 1/6

15/7 : 5/21 = 9

a) \(\left(\dfrac{1}{6}+\dfrac{5}{9}\right)+\dfrac{4}{9}\)

\(=\dfrac{1}{6}+\dfrac{5}{9}+\dfrac{4}{9}\)

\(=\dfrac{1}{6}+1\)

\(=\dfrac{7}{6}\)

b) \(\dfrac{3}{17}+\left(\dfrac{14}{17}-\dfrac{2}{3}\right)\)

\(=\dfrac{3}{17}+\dfrac{14}{17}-\dfrac{2}{3}\)

\(=1-\dfrac{2}{3}\)

\(=\dfrac{1}{3}\)

c) \(\left(\dfrac{3}{2}-\dfrac{2}{3}\right)+\dfrac{7}{6}\)

\(=\left(\dfrac{9}{6}-\dfrac{4}{6}\right)+\dfrac{7}{6}\)

\(=\dfrac{13}{6}+\dfrac{7}{6}\)

\(=\dfrac{20}{6}\)

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)