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a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
a: \(\dfrac{4\cdot5+4\cdot11}{8\cdot7+4\cdot3}=\dfrac{4\cdot16}{4\left(2\cdot7+3\right)}=\dfrac{16}{17}=\dfrac{11088}{11781}\)
\(\dfrac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}=\dfrac{-120+170}{30+60}=\dfrac{50}{90}=\dfrac{5}{9}=\dfrac{6545}{11781}\)
\(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=2\cdot5\cdot\dfrac{1}{77}=\dfrac{10}{77}=\dfrac{1530}{11781}\)
Bài 1:
a: 3/4=54/72
-1/9=-8/72
-5/8=-45/72
b: -1/7=-8/56
-1/-8=1/8=7/56
3/4=42/56
\(a,MSC:170\\ \dfrac{7}{10}=\dfrac{7.17}{17.10}=\dfrac{119}{170}\\ \dfrac{-5}{15}=-\dfrac{1}{5}=\dfrac{1.34}{5.34}=\dfrac{34}{170}\\ \dfrac{3}{17}=\dfrac{3.10}{17.10}=\dfrac{30}{17}\\ b,MSC:525\\ \dfrac{-4}{-75}=\dfrac{4}{75}=\dfrac{4.7}{75.7}=\dfrac{28}{525}\\ \dfrac{-3}{5}=\dfrac{-3.105}{5.105}=\dfrac{-315}{525}\\ \dfrac{8}{35}=\dfrac{8.15}{35.15}=\dfrac{120}{525}\)
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 \(\dfrac{{ - 7}}{{{2^2}.3}} = \dfrac{{ - 7.3}}{{{2^2}{{.3}^2}}} = \dfrac{{ - 21}}{{{2^2}{{.3}^2}}}\)
* Quy tắc nhân 2 phân số: Nhân tử với tử, mẫu với mẫu.
\(\dfrac{8}{3}.\dfrac{3}{7} = \dfrac{{8.3}}{{3.7}} = \dfrac{{24}}{{21}} = \dfrac{{24:3}}{{21:3}} = \dfrac{8}{7}\)
\(\dfrac{4}{6}.\dfrac{5}{8} = \dfrac{{4.5}}{{6.8}} = \dfrac{{20}}{{48}} = \dfrac{{20:4}}{{48:4}} = \dfrac{5}{{12}}\)
1: Rút gọn
a) \(-\dfrac{33}{51}=\dfrac{-33:3}{51:3}=\dfrac{-11}{17}\)
b) \(\dfrac{156}{-168}=\dfrac{-156}{168}=\dfrac{-156:12}{168:12}=\dfrac{-13}{14}\)
c) \(\dfrac{-75}{-100}=\dfrac{75}{100}=\dfrac{75:25}{100:25}=\dfrac{3}{4}\)
2: Quy đồng mẫu:
a) \(-\dfrac{3}{8}\) và \(\dfrac{5}{4}\)
MSC: 8
\(-\dfrac{3}{8}=\dfrac{-3}{8}\)
\(\dfrac{5}{4}=\dfrac{5\cdot2}{4\cdot2}=\dfrac{10}{8}\)
b) \(\dfrac{-7}{6}\) ; \(\dfrac{5}{12}\) và \(-\dfrac{5}{6}\)
MSC: 12
\(-\dfrac{7}{6}=\dfrac{-7\cdot2}{6\cdot2}=\dfrac{-14}{12}\)
\(\dfrac{5}{12}=\dfrac{5}{12}\)
\(\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)
\(\frac{3}{-7}\)= \(\frac{3x8}{-7x8}\)=\(\frac{24}{-56}\)
\(\frac{5}{8}\)=\(\frac{5x-7}{8x-7}\)=\(\frac{-35}{-56}\)
Vậy quy đồng mẫu số \(\frac{3}{-7}\)và \(\frac{5}{8}\)là : \(\frac{24}{-56}\)và \(\frac{-36}{-56}\)
Nếu so sánh thì ta thấy :\(\frac{24}{-56}\)>\(\frac{-36}{-56}\)