\(a,\dfrac{1}{2}:\dfrac{2}{4}:5\) -> chỉ có phép chia nên thực hiện từ trái sang phải :>

\(=1:5=\dfrac{1}{5}\) 

\(b,\dfrac{2}{5}:12:\dfrac{4}{3}\) -> tương tự câu thứ nhất :>

\(=\dfrac{1}{30}:\dfrac{4}{3}=\dfrac{1}{40}\)

1)

\(\dfrac{1}{2}:\dfrac{2}{4}:\dfrac{5}{1}=\left(\dfrac{1}{2}.\dfrac{4}{2}\right).\dfrac{1}{5}=1.\dfrac{1}{5}=\dfrac{1}{5}\)

dấu chấm là dấu nhân á .

2)

\(\dfrac{2}{5}:\dfrac{12}{1}:\dfrac{4}{3}=\left(\dfrac{2}{5}.\dfrac{1}{12}\right).\dfrac{3}{4}=\dfrac{1}{30}.\dfrac{3}{4}=\dfrac{3}{120}=\dfrac{1}{40}\)

`1 2 / 3 : 1 / 2 + 0,8 : 2 / 3`

`= 5 / 3 xx 2 + 4 / 5 xx 3 / 2`

`= [ 5 xx 2 ] / 3 + [ 2 xx 2 xx 3 ] / [ 5 xx 2 ]`

`= 10 / 3 + 6 / 5`

`= 50 / 15 + 18 / 15 = 68 / 15`

=1/10+1/10+...+1/10=10/10=1

=1

a: \(=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}=\dfrac{15}{48}=\dfrac{5}{16}\)

b: \(=\dfrac{2}{3}+\dfrac{12}{42}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\)

c: \(=\dfrac{24}{5}\cdot\dfrac{5}{12}=2\)

d: \(=\dfrac{1}{9}+\dfrac{3}{4}=\dfrac{4+27}{36}=\dfrac{31}{36}\)

1: \(=\dfrac{3}{8}\cdot\dfrac{5}{6}=\dfrac{15}{24}=\dfrac{5}{8}\)

2: \(=\dfrac{2}{3}+\dfrac{12}{42}=\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{14+6}{21}=\dfrac{20}{21}\)

3: \(=\dfrac{24}{5}\cdot\dfrac{5}{12}=2\)

4: \(=\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{4+9}{12}=\dfrac{13}{12}\)

`1/2xx3/4:6/5= 1/2 xx 3/4 xx 5/6 = 3/8 xx 5/6 =5/16`           

`2/3+1/6:7/12 = 2/3 + 1/6 xx 12/7 = 2/3 + 2/7 = 14/21 + 6/21 = 20/21`       

`8xx3/5:12/5= 8xx 3/5 xx 5/12 = 24/5 xx 5/12 = 2`         

`4/9+3/4-1/3= (16/36 + 27/36 ) - 1/3 = 43/36 - 1/3 = 31/36`

`1/2xx3/4:6/5`

`=1/2xx3/4xx5/6`

`=3/8xx5/6`

`=15/48`

`=5/16`

__

`2/3+1/6:7/12`

`=2/3+1/6xx12/7`

`=2/3+2/7`

`=14/21+6/21`

`=20/21`

__

`8xx3/5:12/5`

`=8xx3/5xx5/12`

`=8xx1/4`

`=8/4`

`=2`

__

`4/9+3/4-1/3`

`=(4/9-1/3)+3/4`

`=(4/9-3/9)+3/4`

`=1/9+3/4`

`=4/36+27/36`

`=31/36`

a)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{99.100}\)

=\(9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

=\(9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

=\(9.\left(\frac{1}{1}-\frac{1}{100}\right)\)

=\(9.\frac{99}{100}\)

=\(\frac{891}{100}\)

Cách tìm BCNN:

\(\dfrac{1}{3}\times\dfrac{16}{1}:\dfrac{8}{3}=\left(\dfrac{1}{3}\times\dfrac{16}{1}\right):\dfrac{8}{3}=\dfrac{16}{3}:\dfrac{8}{3}=\dfrac{16}{3}\times\dfrac{3}{8}=\dfrac{48}{24}=2\)

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