a) \(\frac{16}{35}+\frac{8}{35}=\frac{24}{35}\)

b)\(\frac{160}{77}-\frac{28}{77}=\frac{132}{77}=\frac{12}{1}=12\)

c)\(\frac{72}{180}=\frac{18}{45}\)

d) \(\frac{90}{360}=\frac{1}{4}\)

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:

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

\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow A=\frac{8}{99}\)

\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)

\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\frac{32}{99}\)

\(A=\frac{8}{99}\)

\(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}.\)

    \(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(4A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}\)\(=\frac{33-1}{99}\)\(=\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}:4=\frac{32}{99}.\frac{1}{4}=\frac{8}{99}\)

\(a.\) \(\frac{4}{15}.\frac{7}{9}+\frac{4}{15}.\frac{2}{9}\)

\(=\frac{4}{15}\left(\frac{7}{9}+\frac{2}{9}\right)\)

\(=\frac{4}{15}.\frac{9}{9}\)

\(=\frac{4}{15}.1\)

\(=\frac{4}{15}\)

\(b.\) \(\frac{13}{19}.\frac{23}{11}-\frac{13}{19}.\frac{8}{11}-\frac{13}{19}.\frac{4}{11}\)

\(=\frac{13}{19}\left(\frac{23}{11}-\frac{8}{11}-\frac{4}{11}\right)\)

\(=\frac{13}{19}.\frac{11}{11}\)

\(=\frac{13}{19}.1\)

\(=\frac{13}{19}\)

a)4/15 x(7/9+2/9)=4/15x1=4/15

b)13/19x(23/11-8/11-4/11)13/19x1=13/19

A) \(\frac{4}{15}x\frac{7}{9}+\frac{4}{15}x\frac{2}{9}\)

\(=\frac{28}{135}+\frac{8}{135}\)

\(=\frac{36}{135}=\frac{4}{15}\)

B) \(\frac{13}{19}x\frac{23}{11}-\frac{13}{19}x\frac{8}{11}-\frac{13}{19}x\frac{4}{11}\)

\(=\frac{299}{209}-\frac{104}{209}-\frac{52}{209}\)

\(=\frac{143}{209}\)

\(\frac{-4}{7}\times\frac{5}{11}+\frac{-4}{7}\times\frac{6}{11}+\)\(2017\frac{4}{7}\)

\(=\frac{-4}{7}\times\left(\frac{5}{11}+\frac{6}{11}\right)+2017+\frac{4}{7}\)

\(=\frac{-4}{7}\times1+\frac{4}{7}+2017\)

\(=\left(\frac{-4}{7}+\frac{4}{7}\right)+2017\)

\(=0+2017=2017\)

1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)

2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)

Vậy ......

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