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\(a,MSC:170\\ \dfrac{7}{10}=\dfrac{7.17}{17.10}=\dfrac{119}{170}\\ \dfrac{-5}{15}=-\dfrac{1}{5}=\dfrac{1.34}{5.34}=\dfrac{34}{170}\\ \dfrac{3}{17}=\dfrac{3.10}{17.10}=\dfrac{30}{17}\\ b,MSC:525\\ \dfrac{-4}{-75}=\dfrac{4}{75}=\dfrac{4.7}{75.7}=\dfrac{28}{525}\\ \dfrac{-3}{5}=\dfrac{-3.105}{5.105}=\dfrac{-315}{525}\\ \dfrac{8}{35}=\dfrac{8.15}{35.15}=\dfrac{120}{525}\)
MSC : 360
`5/8 = (5xx45)/(8xx45)=225/360`
`7/20=(7xx18)/(20xx18)=126/360`
`-4/9=(-4xx40)/(9xx40)=-160/360`
\(\dfrac{5}{7}=\dfrac{75}{105};-\dfrac{3}{21}=-\dfrac{1}{7}=\dfrac{-15}{105};\dfrac{-8}{15}=\dfrac{-56}{105}\)
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, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} = \dfrac{-1}{{28}}\)
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
a: 5/14=10/28
-3/7=-12/28
-3/4=-21/28
b: 4/11=40/110
-3/5=-66/110
-1/-2=1/2=55/110
c: 7/15=56/120
-3/8=-45/120
-2/3=-80/120
