mik cần rất rất gấp ạ

Bài 2: 

a: 1/3=20/60

1/5=12/60

1/12=5/60

b: 1/3=16/48

1/12=4/48

1/48=1/48

Bài 2: 

a: 1/3=20/60

1/5=12/60

1/12=5/60

b: 1/3=16/48

1/12=4/48

1/48=1/48

Bài 2: 

a: 1/3=20/60

1/5=12/60

1/12=5/60

b: 1/3=16/48

1/12=4/48

1/48=1/48

Bài 2: 

a: 1/3=20/60

1/5=12/60

1/12=5/60

b: 1/3=16/48

1/12=4/48

1/48=1/48

Bài 2: 

a: 1/3=20/60

1/5=12/60

1/12=5/60

b: 1/3=16/48

1/12=4/48

1/48=1/48

`a,MSC:30`

`1/2=(1xx15)/(2xx15)=15/30`

`2/3=(2xx10)/(3xx10)= 20/30`

`3/5=(3xx6)/(5xx6)=18/30`

`b,MSC:12`

`2/3=(2xx4)/(3xx4)=8/12`

`3/4=(3xx3)/(4xx3)=9/12`

`7/12` giữ nguyên

\(a,\dfrac{1}{2}=\dfrac{1\times15}{2\times15}=\dfrac{15}{30}\\ \dfrac{2}{3}=\dfrac{2\times10}{3\times10}=\dfrac{20}{30}\\ \dfrac{3}{5}=\dfrac{3\times6}{5\times6}=\dfrac{18}{30}\\ b,\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\\ \dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\\ \dfrac{7}{12}giữnguyên\)

phân tích ra thừa số nguyên tố

`5=5`

`12=2^2 xx3`

`=>` MSC nhỏ nhất là: `5xx12=60`

`2/5=(2xx12)/(5xx12)=24/60`

`5/12=(5xx5)/(12xx5)=25/60`

\(\dfrac{2}{5};\dfrac{5}{12}\left(MSC=60\right)\)

\(\dfrac{2}{5}=\dfrac{2\times12}{5\times12}=\dfrac{24}{60};\dfrac{5}{12}=\dfrac{5\times5}{12\times5}=\dfrac{25}{60}\)

`MSC: 60`

\(\dfrac{2}{5}=\dfrac{2\times12}{5\times12}=\dfrac{24}{60};\dfrac{5}{12}=\dfrac{5\times5}{12\times5}=\dfrac{25}{60}\)

a: 1/54=1/54

5/6=45/54

b: 3/22=3/22

6/11=12/22

c: 1/2=6/12

2/3=8/12

3/4=9/12

xem trong SGK

\(\dfrac{7\times2}{1\times2}=\dfrac{14}{2},\dfrac{9}{2};\dfrac{81\times3}{7\times3}=\dfrac{243}{21},\dfrac{9}{21}\\ \dfrac{4\times2}{3\times2}=\dfrac{8}{6},\dfrac{16}{6};\dfrac{5\times3}{2\times3}=\dfrac{15}{6},\dfrac{19\times2}{3\times2}=\dfrac{38}{6}\)

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 \(\dfrac{{ - 7}}{{{2^2}.3}} = \dfrac{{ - 7.3}}{{{2^2}{{.3}^2}}} = \dfrac{{ - 21}}{{{2^2}{{.3}^2}}}\)

1. a) Ta có BCNN(12, 15) = 60 nên ta lấy mẫu chung của hai phân số là 60. 

Thừa số phụ:

60:12 =5; 60:15=4

Ta được:

\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\)

\(\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\)

 b) Ta có BCNN(7, 9, 12) = 252 nên ta lấy mẫu chung của ba phân số là 252. 

Thừa số phụ:

252:7 = 36; 252:9 = 28; 252:12 = 21

Ta được:

\(\frac{2}{7} = \frac{{2.36}}{{7.36}} = \frac{{72}}{{252}}\)

\(\frac{4}{9} = \frac{{4.28}}{{9.28}} = \frac{{112}}{{252}}\)

\(\frac{7}{{12}} = \frac{{7.21}}{{12.21}} = \frac{{147}}{{252}}\)

2. a) Ta có BCNN(8, 24) = 24 nên:

\(\frac{3}{8} + \frac{5}{{24}} = \frac{{3.3}}{{8.3}} + \frac{5}{{24}} = \frac{9}{{24}} + \frac{5}{{24}} = \frac{{14}}{{24}} = \frac{7}{{12}}\)

 b) Ta có BCNN(12, 16) = 48 nên:

\(\frac{7}{{16}} - \frac{5}{{12}} = \frac{{7.3}}{{16.3}} - \frac{{5.4}}{{12.4}} = \frac{{21}}{{48}} - \frac{{20}}{{48}} = \frac{1}{{48}}\).

a: 3/4=135/180

-1/9=-20/180

-5/5=-1=-180/180

1.3:

a: \(\dfrac{5}{14}=\dfrac{5\cdot3}{14\cdot3}=\dfrac{15}{42}\)

\(\dfrac{4}{21}=\dfrac{4\cdot2}{21\cdot2}=\dfrac{8}{42}\)

b: \(\dfrac{4}{5}=\dfrac{4\cdot12}{5\cdot12}=\dfrac{48}{60}\)

\(\dfrac{7}{12}=\dfrac{7\cdot5}{12\cdot5}=\dfrac{35}{60}\)

\(\dfrac{8}{5}=\dfrac{8\cdot12}{5\cdot12}=\dfrac{96}{60}\)

1.2:

a: \(21=3\cdot7;36=3^2\cdot2^2\)

=>\(ƯCLN\left(21;36\right)=3>1\)
=>Phân số này chưa tối giản

\(\dfrac{21}{36}=\dfrac{21:3}{36:3}=\dfrac{7}{12}\)

b: \(23=23;73=73\)

=>\(ƯCLN\left(23;73\right)=1\)

=>23/73 là phân số tối giản

1.1:

theo đề ta có: 480⋮a và 720⋮a

=> a = ƯCLN(480,720)

480=2 mũ 5.3.5

720=2 mũ 4.3 mũ 2.5

=> ƯCLN(420,720)= 2 mũ 4.3.5=240

=> a=240