Ta có: \(\dfrac{5}{7} = \dfrac{{5.4}}{{7.4}} = \dfrac{{20}}{{28}}\) và \(\dfrac{{ - 3}}{4} = \dfrac{{ - 3.7}}{{4.7}} = \dfrac{{ - 21}}{{28}}\)

Như vậy, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} =  \dfrac{-1}{{28}}\)

a) \(\dfrac{3}{5}=\dfrac{21}{35};\dfrac{4}{7}=\dfrac{20}{35}\)

b) \(\dfrac{5}{6}=\dfrac{45}{54};\dfrac{7}{9}=\dfrac{42}{54}\)

Lời giải:

a. $\frac{5}{9}=\frac{5\times 2}{9\times 2}=\frac{10}{18}$
b. $\frac{9}{20}=\frac{9\times 3}{20\times 3}=\frac{27}{60}$

a) \(\dfrac{5}{9}=\dfrac{5\times2}{9\times2}=\dfrac{10}{18}\)

b) \(\dfrac{9}{20}=\dfrac{9\times3}{20\times3}=\dfrac{27}{60}\)

a: 7/15=56/120

-3/8=-45/120

-2/3=-80/120

a) \(\dfrac{1}{6}=\dfrac{3}{18}\)

b) \(\dfrac{4}{5}=\dfrac{48}{60}\)

c) \(\dfrac{7}{25}=\dfrac{28}{100}\)

21/35 

20/35

3/5=21/35

4/7=20/35

ta có : \(BCNN\left(7;21;15\right)=105\\ \dfrac{5}{7}=\dfrac{75}{105};\dfrac{-3}{21}=\dfrac{-15}{105};\dfrac{-8}{15}=\dfrac{-56}{105}\)

 7 = 7;       21 = 3. 7;            15 = 3. 5

Mẫu chung: BCNN(7; 21; 15) = 3. 5. 7 = 105

Thừa số phụ: 105: 7 = 15; 105: 21 = 5; 105: 15 = 7

\(\dfrac{5}{7}=\dfrac{5.15}{7.15}=\dfrac{75}{105}\)

\(\dfrac{-3}{21}=\dfrac{-3.5}{21.5}\dfrac{-15}{105}\)

\(\dfrac{-8}{15}=\dfrac{-8.7}{15.7}=\dfrac{-56}{105}\)

\(\dfrac{5}{7}=\dfrac{75}{105}\)

\(\dfrac{-3}{21}=\dfrac{-15}{105}\)

\(\dfrac{-8}{15}=\dfrac{-56}{105}\)

5/7 =75/105

-3/21= -15/105

-8/15=-56/105

2/5 và 5/7 

\(\dfrac{2}{5}=\dfrac{2\times7}{5\times7}=\dfrac{14}{35}\)

\(\dfrac{5}{7}=\dfrac{5\times5}{7\times5}=\dfrac{25}{35}\)

x 1 à bn

a) \(\dfrac{2}{3}=\dfrac{2.8}{3.8}=\dfrac{16}{24}\)

\(\dfrac{5}{8}=\dfrac{5.3}{8.3}=\dfrac{15}{24}\)

b) \(\dfrac{1}{4}=\dfrac{1.3}{4.3}=\dfrac{3}{12}\)

\(\dfrac{7}{12}=\dfrac{7.1}{12.1}=\dfrac{7}{12}\)

c) \(\dfrac{5}{6}=\dfrac{5.4}{6.4}=\dfrac{20}{24}\)

\(\dfrac{3}{8}=\dfrac{3.3}{8.3}=\dfrac{9}{24}\)

a) 

b) 

c)