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:

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

\(A=\frac{17}{23}\cdot\frac{8}{16}\cdot\frac{23}{17}\cdot\left(-80\right)\cdot\frac{3}{4}\)\(=\frac{17\cdot4\cdot2\cdot23\cdot16\cdot\left(-5\right)\cdot3}{23\cdot16\cdot17\cdot4}\)

=> \(A=\frac{2\cdot\left(-5\right)\cdot3}{1}=-30\)

\(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)

=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{7}{12}-\frac{7}{12}\right)\)

=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\cdot0=0\)

a)\(A=\frac{17}{23}.\frac{8}{16}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)

    \(A=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{8}{16}.\frac{3}{4}\right).\left(-80\right)\)

     \(A=\frac{3}{8}.\left(-80\right)\)

     \(A=-30\)

b)\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)

   \(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)

   \(C=0\)

\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)

\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)

\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)

\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)

\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)

Đến đây là tính dễ rồi :v

\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)

\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)

\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)

\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)

Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)

\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)