\(A=\frac{n+6}{n-1}=\frac{n-1+7}{n-1}=1+\frac{7}{n-1}\inℤ\Leftrightarrow\frac{7}{n-1}\inℤ\)

mà \(n\)là số nguyên nên \(n-1\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\)

\(\Leftrightarrow n\in\left\{-6,0,2,8\right\}\).

31.(65+35)-500

31.100-500

3100-500

2600

\(31.65+31.35-500\)

\(=32.(65+35)-500\)

\(=32.100-500\)

\(=3200-500\)

\(=2700\)

đè ngị bạn viết lại đề

\(\left(x-1\right)^5=-32\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)

Suy ra: \(x-1=-2\Leftrightarrow x=-1\).

( x - 1 )⁵ = ( - 2 )⁵ => x - 1 = - 2 => x =( - 2 ) + 1 => x = -1    

\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\)

\(=\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{9.11}{10.10}\)

\(=\frac{\left(1.3\right).\left(2.4\right).....\left(9.11\right)}{\left(2.2\right).\left(3.3\right).....\left(10.10\right)}\)

\(=\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}\)

\(=\frac{1.11}{10.2}=\frac{11}{20}\)

\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)

\(\Rightarrow S=\frac{3}{4}-\frac{1}{4}-\left(\frac{7}{3}-\frac{9}{2}\right)-\frac{5}{6}\)

\(\Rightarrow S=\frac{1}{2}-\frac{7}{3}+\frac{9}{2}-\frac{5}{6}\)

\(\Rightarrow S=\frac{3}{6}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)

\(\Rightarrow S=\frac{11}{6}\)