Câu 1:\(\frac{12}{23}\times\frac{9}{28}+\frac{11}{23}\times\frac{9}{28}\)

\(=\frac{9}{28}\times\left(\frac{12}{23}+\frac{11}{23}\right)\)

\(=\frac{9}{28}\times\frac{23}{23}\)

\(=\frac{9}{28}\times1\)

\(=\frac{9}{28}\)

Câu 2:

\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)

\(=\left(3+2+2+3\right)+\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)\)

\(=10+\frac{9}{9}\)

\(=10+1\)

\(=11\)

\(\frac{12}{23}x\frac{9}{28}+\frac{11}{23}x\frac{9}{28}=\)\(\left(\frac{12}{23}+\frac{11}{23}\right)x\frac{9}{28}\)\(=\) \(1x\frac{9}{28}=\frac{9}{28}\)

\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)\(=\) \(\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)+\left(3+2+3+2\right)\)\(=\) \(1+3+2+3+2=\)\(11\) 

\(\frac{12}{23}\times\frac{9}{28}+\frac{13}{23}\times\frac{9}{28}=\left(\frac{12}{23}+\frac{13}{23}\right)\times\frac{9}{28}\)

                                         \(=\) \(\frac{25}{23}\times\frac{9}{28}\)

                                         \(=\frac{225}{664}\)

a) \(\left(\frac{3}{5}.\frac{5}{3}\right).\frac{8}{27}\)

=  \(1.\frac{8}{27}\)

=\(\frac{8}{27}\)

b) \(\frac{7}{19}.\left(\frac{1}{3}.\frac{2}{3}\right)\)

= \(\frac{7}{19}.\frac{2}{9}\)

=\(\frac{14}{81}\)

Không biết có đúng không? Tick mình nha !

\(\frac{7}{19}\)x\(\frac{1}{3}\)+\(\frac{7}{19}\)x\(\frac{2}{3}\)=\(\frac{7}{19}\)x(\(\frac{1}{3}\)+\(\frac{2}{3}\))=\(\frac{7}{19}\)x1=\(\frac{7}{19}\)

\(\frac{7}{19}\)x \(\frac{1}{3}\)x \(\frac{7}{19}\)x \(\frac{2}{3}\)

= \(\frac{7}{19}\)x (\(\frac{1}{3}\)+ \(\frac{2}{3}\))

= \(\frac{7}{19}\)x 1

= \(\frac{7}{19}\)

Chúc bạn học tốt!

a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)

b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0

a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}\)

\(=\frac{25}{33}\)

b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)

Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.

Ta có \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

 = \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

= \(\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

=     \(\frac{2}{3}+\frac{1}{11}\)

=       \(\frac{25}{33}\)

\(\frac{3}{5}.\frac{8}{27}.\frac{5}{3}=1.\frac{8}{27}.1=\frac{8}{27}\)

\(\frac{7}{19}.\frac{1}{3}+\frac{7}{19}.\frac{2}{3}=\frac{7}{19}.\left(\frac{1}{3}+\frac{2}{3}\right)=\frac{7}{19}.1=\frac{7}{19}\)

\(\frac{12}{5}.4-4.\frac{7}{5}=4\left(\frac{12}{5}-\frac{7}{5}\right)=4.1=4\)

\(\frac{3}{5}x\frac{8}{27}x\frac{5}{3}\)

\(=\frac{3}{5}x\frac{5}{3}x\frac{8}{27}\)

\(=1x\frac{8}{27}\)

\(=\frac{8}{27}\)

\(\frac{7}{19}x\frac{1}{3}+\frac{7}{19}x\frac{2}{3}\)

\(=\frac{7}{19}x\left(\frac{1}{3}+\frac{2}{3}\right)\)

\(=\frac{7}{19}x1=\frac{7}{19}\)

\(\frac{12}{5}x4-4x\frac{7}{5}\)

\(=4x\left(\frac{12}{5}-\frac{7}{5}\right)\)

\(=4x1=4\)

Đúng luôn nên các bn nhớ k mk nhé

A) 3/5 x 8/27 x 5/3

= ( 3/5 x 8/27 ) x 5/3

= 8/45 x 5/3 

= 8/27

B) 7/19 x 1/3 + 7/19 x 2/3

= ( 7/19 x 1/3 ) + ( 7/19 x 2/3 )

= 7/57 + 14/57

= 7/19

C) 12/5 x 4 - 4 x 7/5

= ( 12/5 x 4 ) - ( 4 x 7/5 )

= 48/5 - 28/5

= 4

Để nhân các phân số này, ta chỉ cần nhân tử số với nhau và mẫu số với nhau:

\[
\frac{1}{3} \times \frac{2}{5} \times \frac{3}{7} \times \frac{4}{9} \times \frac{5}{11} \times \frac{6}{15} \times \frac{7}{15} \times \frac{8}{15} \times \frac{9}{19} \times \frac{10}{21} \times \frac{11}{32} \times \frac{12}{25} \times \left( \frac{126}{252} - 4 \right)
\]

Sau đó, ta thực hiện các phép tính:

1. Nhân tử số:
\[1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12 \times 126 = 997920\]

2. Nhân mẫu số:
\[3 \times 5 \times 7 \times 9 \times 11 \times 15 \times 15 \times 15 \times 19 \times 21 \times 32 \times 25 \times 252 = 7621237680\]

Kết quả là:
\[\frac{997920}{7621237680}\]

Bây giờ, ta có thể rút gọn phân số này bằng cách chia tử số và mẫu số cho 160:

\[ \frac{997920}{7621237680} = \frac{997920 ÷ 160}{7621237680 ÷ 160} = \frac{6237}{47695230} \]

a) $\frac{1}{6} \times \frac{2}{3} = \frac{{1 \times 2}}{{6 \times 3}} = \frac{2}{{18}} = \frac{1}{9}$                            

b) $\frac{6}{5} \times \frac{3}{8} = \frac{{6 \times 3}}{{5 \times 8}} = \frac{{18}}{{40}} = \frac{9}{{20}}$

c) $\frac{4}{3} \times \frac{8}{9} = \frac{{4 \times 8}}{{3 \times 9}} = \frac{{32}}{{27}}$        

d) $\frac{5}{{12}} \times \frac{{12}}{5} = \frac{{5 \times 12}}{{12 \times 5}} = \frac{{60}}{{60}} = 1$

bài 1:

\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)

\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)

\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)

bài 2:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)

\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)

bài 3: 

a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)

b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)

bài 4: 

a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)

b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)

Bài 1: 

  6/11 + 1/3 + 5/11 

= ( 6/11 + 5/11) + 1/3

= 11/11 + 1/3

= 1 + 1/3 

= 3/3 +1/3  

= 4/3 

Bài 2: 

1/2 + 1/6 + 1/12 + 1/20 

= ( 1/2 + 1/6 + 1/12 ) + 1/20 

= ( 6/12 + 2/12 + 1/12 ) + 1/20 

=9/12 + 1/20 

= 3/4 +1/20  

= 15/20 + 1/20 

= 16/20 = 4/5

 Bài 3:

a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\) 

b)  \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)   

Bài 4:

a) 2/5  x 1/4 + 3/4 x 2/5 

= 2/5 x ( 1/4  + 3/4) 

 = 2/5 x  1 

= 2/5

 b) 6/11 : 2/3 +5/11 : 2/3 

=  ( 6/11 + 5/11) x 3/2 

= 11/11 x 3/2  

= 1 x 3/2  

=  3/2  

....  

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)