Quy đồng mẫu các phân số sau:
\(\dfrac{5}{7};\dfrac{-3}{21};\dfrac{-8}{15}\).
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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, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} = \dfrac{-1}{{28}}\)
2/5 và 5/7
\(\dfrac{2}{5}=\dfrac{2\times7}{5\times7}=\dfrac{14}{35}\)
\(\dfrac{5}{7}=\dfrac{5\times5}{7\times5}=\dfrac{25}{35}\)
Lời giải:
a. $\frac{5}{9}=\frac{5\times 2}{9\times 2}=\frac{10}{18}$
b. $\frac{9}{20}=\frac{9\times 3}{20\times 3}=\frac{27}{60}$
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
\(\dfrac{5}{7}=\dfrac{5.15}{7.15}=\dfrac{75}{105}\)
\(\dfrac{-3}{21}=\dfrac{-3.5}{21.5}\dfrac{-15}{105}\)
\(\dfrac{-8}{15}=\dfrac{-8.7}{15.7}=\dfrac{-56}{105}\)
a) Ta có BCNN(3,7)=21
Thừa số phụ: 21:3=7 và 21:7=3
\(\dfrac{2}{3} = \dfrac{{2.7}}{{3.7}} = \dfrac{{14}}{{21}}\) và \(\dfrac{{ - 6}}{7} = \dfrac{{ - 6.3}}{{7.3}} = \dfrac{{ - 18}}{{21}}\)
b) Ta có \(BCNN\left( {\left( {{2^2}{{.3}^2}} \right),\left( {{2^2}.3} \right)} \right) = {2^2}{.3^2}\)
Thừa số phụ \(\left( {{2^2}{{.3}^2}} \right):\left( {{2^2}.3^2} \right) = 1\) và \(\left( {{2^2}{{.3}^2}} \right):\left( {{2^2}.3} \right) = 3\)
\(\dfrac{5}{{{2^2}{{.3}^2}}}\) và \(\dfrac{{ - 7}}{{{2^2}.3}} = \dfrac{{ - 7.3}}{{{2^2}{{.3}^2}}} = \dfrac{{ - 21}}{{{2^2}{{.3}^2}}}\)
a: \(\dfrac{5}{7}=\dfrac{5\cdot11}{7\cdot11}=\dfrac{55}{77}\)
\(\dfrac{9}{11}=\dfrac{9\cdot7}{11\cdot7}=\dfrac{63}{77}\)
b: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot9}{7\cdot9}=\dfrac{54}{63}\)
\(-\dfrac{12}{54}=\dfrac{-2}{9}=\dfrac{-2\cdot7}{9\cdot7}=-\dfrac{14}{63}\)
c: \(\dfrac{-11}{30}=\dfrac{-11\cdot4}{30\cdot4}=\dfrac{-44}{120}\)
\(\dfrac{-17}{-40}=\dfrac{17}{40}=\dfrac{17\cdot3}{40\cdot3}=\dfrac{51}{120}\)
d: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot3}{7\cdot3}=\dfrac{18}{21}\)
\(\dfrac{-12}{36}=\dfrac{-1}{3}=\dfrac{-1\cdot7}{3\cdot7}=\dfrac{-7}{21}\)
\(\dfrac{5}{7}=\dfrac{75}{105}\)
\(\dfrac{-3}{21}=\dfrac{-15}{105}\)
\(\dfrac{-8}{15}=\dfrac{-56}{105}\)
5/7 =75/105
-3/21= -15/105
-8/15=-56/105