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) Mẫu số chung = BCNN(11, 7) = 77

Thừa số phụ: 77: 11= 7; 77:7 = 11.

Ta có:

\(\begin{array}{l}\frac{7}{{11}} + \frac{5}{7} = \frac{{7.7}}{{11.7}} + \frac{{5.11}}{{7.11}}\\ = \frac{{49}}{{77}} + \frac{{55}}{{77}} = \frac{{104}}{{77}}\end{array}\).

b) Mẫu số chung = BCNN(20, 15)= 60 

Thừa số phụ: 60:20 = 3; 60:15 = 4

Ta có:

\(\begin{array}{l}\frac{7}{{20}} - \frac{2}{{15}} = \frac{{7.3}}{{20.3}} - \frac{{2.4}}{{15.4}}\\ = \frac{{21}}{{60}} - \frac{8}{{60}} = \frac{{13}}{{60}}\end{array}\).

a)\(\frac{1}{8}.\frac{3}{5} = \frac{{1.3}}{{8.5}} = \frac{3}{{40}}\)           

b)\(\frac{{ - 6}}{7}:\left( { - \frac{5}{3}} \right) = \frac{{ - 6}}{7}.\frac{{ - 3}}{5} = \frac{{18}}{{35}}\)           

c)\(0,6.\left( { - 0,15} \right) = \frac{6}{{10}}.\frac{{ - 15}}{{100}} = \frac{{ - 90}}{{1000}} = \frac{{ - 9}}{{100}}\).

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

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,5/9+(-3/5)  +( -5/9)

=5/9+(-5/9)+(-3/5)

=0+(-3/5)

=-3/5

b, b,-3/7. 2/11- (-3/7).9/11

=-3/7.(2/11-9/11)

=-3/7.(-7/11)

=3/11

c,,2/3 + 5/6 : 5 - 1/18 . (-3)

=2/3+(1/6-(-1/6))

=2/3+1/3

=3/3

=1

tick nếu đúng nha

Dễ mà, dùng máy tính là xong

= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)

= \(\frac{2}{7}.2\)

= \(\frac{4}{7}\)

    \(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)

=> \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)

=> \(\frac{2}{7}.2\)

=> \(\frac{4}{7}\)

#Hk_tốt

#Ngọc's_Ken'z

\(\begin{array}{l}a)0,36.\frac{{ - 5}}{9}\\ = \frac{{36}}{{100}}.\frac{{ - 5}}{9}\\ = \frac{9}{{25}}.\frac{{ - 5}}{9}\\ = \frac{{ - 1}}{5}\\b)\frac{{ - 7}}{6}:1\frac{5}{7}\\ = \frac{{ - 7}}{6}:\frac{{12}}{7}\\ = \frac{{ - 7}}{6}.\frac{7}{{12}}\\ = \frac{{ - 49}}{{72}}\end{array}\)

Chú ý: Khi tính toán, nếu phân số chưa ở dạng tối giản thì ta nên rút gọn về dạng tối giản để tính toán thuận tiện hơn.

a)-2

b)-7

c)12

e)4

c)-15

g)8

a) -2
b) -7
c) 12
e) 4
c) -15
g) 8