b, 

Để \(\dfrac{2n+7}{n-2}\) nhận giá trị nguyên . 

\(\rightarrow 2n + 7 \vdots n-2\)

\(\rightarrow 2n -4 +11 \vdots n-2\)

\(\rightarrow 2( n-2 ) +11 \vdots n-2\)

Do \(2n -4 \vdots n-2\) mà để \(2n + 7 \vdots n-2 .\)

\(\rightarrow 11 \vdots n-2\)

\(\rightarrow n-2 \in Ư_{11} =\) { \(\pm 1 ; \pm11\) }

Lập bảng :

\(\begin{array}{|c|c|c|}\hline \text{n-2}&\text{1}&\text{-1}&\text{11}&\text{-11}\\\hline \text{n}&\text{3}&\text{1}&\text{13}&\text{-9}\\\ \\\hline\end{array}\)

Vậy \(x \in\) { \(3;1;13;-9\) }

a,

Để \(\dfrac{2n+7}{n-2}\) là phân số .

\(\rightarrow n-2 \ne 0\) 

\(\rightarrow n \ne 2\)

Vậy \(n \ne 2 \) thì \(\dfrac{2n+7}{n-2}\) là phân số .

\(\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9}{15}+\dfrac{10}{15} = \dfrac{19}{15}\)

\(\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{10}{14}-\dfrac{7}{14}=\dfrac{3}{14}\)

\(\dfrac{5}{6} \times \dfrac{4}{5} = \dfrac{4}{6}= \dfrac{2}{3}\)

\( \dfrac{7}{9} : \dfrac{4}{3} = \dfrac{7}{9} \times \dfrac{3}{4} = \dfrac{7}{12}\)

hay đấy 

\(2020^0 = 0\)

\(2011 – [39 – (2^3.3 – 21)^2] : (-3)+ 2021^0\)

\(= 2011 - [ 39 - ( 8.3 - 21 )^2 ] : ( -3 ) + 1\)

\(= 2011 - [ 39 - 3^2 ] : ( -3 ) + 1\)

\(= 2011 - [ 39 - 9 ] : ( -3 ) + 1\)

\(= 2011 - 30 : ( -3 ) + 1\)

\(= 2011 - (-10) + 1\)

\(= 2011 + 10 + 1 = 2022\)

\(-0,305\) và \(-0,36\)

Do \(0,305 < 0,36\)

\(=> -0,305 > -0,36\)

Vậy \(-0,305 > -0,36\)

\(\dfrac{-3}{20}\) và \(\dfrac{2}{-15}\)

Ta có :

\(\dfrac{-3}{20} = \dfrac{-3.3}{20.3}=\dfrac{-9}{60}\)

\(\dfrac{2}{-15} = \dfrac{2.4}{-15.4}= \dfrac{8}{-60}= \dfrac{-8}{60}\)

Do \(-9 <-8\)

\(=> \dfrac{-3}{20} < \dfrac{-2}{15}\)

Vậy \( \dfrac{-3}{20} < \dfrac{-2}{15}\)

\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)

Ta có \(:\)

\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)

\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)

Do \(55 < 60\)

\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)

Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)