a: 

b: 

c:

\(A=\left(\dfrac{1}{49}-\dfrac{1}{2^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{7^2}\right)\left(\dfrac{1}{49}-\dfrac{1}{2^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\left(\dfrac{1}{49}-\dfrac{1}{4}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{10000}\right)\)

=0

a) \(\dfrac{4}{5}\times\dfrac{11}{7}-\dfrac{4}{5}\times\dfrac{1}{7}\)

\(=\dfrac{4}{5}\times\left(\dfrac{11}{7}-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{5}\times\dfrac{10}{7}\)

\(=\dfrac{4}{1}\times\dfrac{2}{7}\)

\(=\dfrac{8}{7}\)

b) \(\dfrac{7}{13}\times\dfrac{25}{6}-\dfrac{7}{13}\times\dfrac{-1}{6}\)

\(=\dfrac{7}{13}\times\left(\dfrac{25}{6}-\dfrac{-1}{6}\right)\)

\(=\dfrac{7}{13}\times\dfrac{26}{6}\)

\(=\dfrac{7}{1}\times\dfrac{2}{6}\)

\(=\dfrac{14}{6}=\dfrac{7}{3}\)

c) \(\dfrac{4}{13}\times\dfrac{5}{11}+\dfrac{4}{13}\times\dfrac{6}{11}-\dfrac{4}{13}\)

\(=\dfrac{4}{13}\times\left(\dfrac{5}{11}+\dfrac{6}{11}\right)-\dfrac{4}{13}\)

\(=\dfrac{4}{13}\times\dfrac{11}{11}-\dfrac{4}{13}\)

\(=\dfrac{4}{13}\times\dfrac{11}{11}-\dfrac{4}{13}\times\dfrac{11}{11}\)

\(=\dfrac{4}{3}\times\left(\dfrac{11}{11}-\dfrac{11}{11}\right)\)

\(=\dfrac{4}{3}\times0=0\)

a)\(\dfrac{4}{5}\cdot\dfrac{11}{7}-\dfrac{4}{5}\cdot\dfrac{1}{7}=\dfrac{4}{5}.\left(\dfrac{11}{7}-\dfrac{1}{7}\right)=\dfrac{4}{5}.\left(\dfrac{11-1}{7}\right)=\dfrac{4}{5}\cdot\dfrac{10}{7}=4\cdot\dfrac{2}{7}=\dfrac{8}{7}\)

b)\(\dfrac{7}{13}\cdot\dfrac{25}{6}-\dfrac{7}{13}\cdot\dfrac{-1}{6}=\dfrac{7}{13}\cdot\left(\dfrac{25}{6}-\dfrac{-1}{6}\right)=\dfrac{7}{13}\cdot\left(\dfrac{25}{6}+\dfrac{1}{6}\right)=\dfrac{7}{13}\cdot\dfrac{26}{6}=\dfrac{7}{13}\cdot\dfrac{13}{3}=\dfrac{7}{3}\)

c)

\(\dfrac{4}{13}\cdot\dfrac{5}{11}+\dfrac{4}{13}\cdot\dfrac{6}{11}-\dfrac{4}{13}=\dfrac{4}{13}.\left(\dfrac{5}{11}+\dfrac{6}{11}-1\right)=\dfrac{4}{13}\cdot\left(1-1\right)=\dfrac{4}{13}\cdot0=0\)

a; \(\dfrac{4}{5}\) \(\times\) \(\dfrac{11}{\dfrac{-4}{5}\times\dfrac{1}{7}}\) = \(\dfrac{-11}{\dfrac{1}{7}}\) = -77

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Có chứ

Để:

\(A\inℤ\)

\(\Leftrightarrow n+8⋮2n-4\)

\(n+8⋮2\left(n-2\right)\)

\(n+8⋮n+2\)

\(n-2+10⋮n-2\)

\(10⋮n-2\)

\(\Rightarrow n-2\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

\(\Rightarrow n\in\left\{3;1;4;0;7;-3;12;-8\right\}\)

Vậy \(\Rightarrow n\in\left\{3;1;4;0;7;-3;12;-8\right\}\) để \(A\) nguyên.

\(\dfrac{-35}{7}< x\le-1\)

\(-35:7< x\le-1\)

\(-5< x< -1\)

Mà \(x\inℤ\)

\(\Rightarrow x\in\left\{-4;-3;-2\right\}\)

1;  \(\dfrac{7}{15}\) + \(\dfrac{8}{15}\) = \(\dfrac{7+8}{15}\) = \(\dfrac{15}{15}\) = 1

2; \(\dfrac{1}{2}\) - \(\dfrac{1}{14}\) = \(\dfrac{1.7}{2.7}\) - \(\dfrac{1}{14}\) = \(\dfrac{7-1}{14}\) = \(\dfrac{6}{14}\) = \(\dfrac{3}{7}\)

3; \(\dfrac{8}{28}\) + \(\dfrac{-21}{35}\) = \(\dfrac{2}{7}\) + \(\dfrac{-21}{35}\)= \(\dfrac{10}{35}\) + \(\dfrac{-21}{35}\) = \(\dfrac{-11}{35}\)

4; \(\dfrac{3}{4}\) + \(\dfrac{2}{3}\) - \(\dfrac{9}{6}\)  = \(\dfrac{9}{12}\) + \(\dfrac{8}{12}\) - \(\dfrac{18}{12}\) = \(\dfrac{9+8-18}{12}\) = \(\dfrac{-1}{12}\)

5; \(\dfrac{11}{36}\)- \(\dfrac{-7}{-24}\) = \(\dfrac{22}{72}\) + \(\dfrac{21}{72}\) = \(\dfrac{53}{72}\)

6; \(\dfrac{4}{15}\) + \(\dfrac{9}{5}\) - \(\dfrac{7}{3}\) = \(\dfrac{4}{15}\) + \(\dfrac{27}{15}\) - \(\dfrac{35}{15}\) = \(\dfrac{-4}{15}\) 

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