\(1+\frac{-1}{60}+\frac{19}{120}

Bài 1 :

\(A=\dfrac{n+1}{n+2}\) có giá trị nguyên âm, dương khi

\(n+1⋮n+2\)

\(\Rightarrow n+1-\left(n+2\right)⋮n+2\)

\(\Rightarrow n+1-n-2⋮n+2\)

\(\Rightarrow-1⋮n+2\)

\(\Rightarrow n+2\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{-3;-1\right\}\left(n\in Z\right)\)

Bài 2 :

\(1+\left(-\dfrac{1}{60}\right)+\dfrac{19}{120}< \dfrac{x}{36}+\left(-\dfrac{1}{60}\right)< \dfrac{58}{90}+\dfrac{59}{72}+\left(-\dfrac{1}{60}\right)\)

\(\Rightarrow1+\dfrac{19}{120}< \dfrac{x}{36}< \dfrac{58}{90}+\dfrac{59}{72}\)

\(\Rightarrow\dfrac{139}{120}< \dfrac{x}{36}< \dfrac{232}{360}+\dfrac{295}{360}\)

\(\Rightarrow\dfrac{417}{360}< \dfrac{10x}{360}< \dfrac{527}{360}\)

\(\Rightarrow417< 10x< 527\)

\(\Rightarrow10x\in\left\{420;430;440;450;460;470;480;490;500;510;520\right\}\)

\(\Rightarrow x\in\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)

Ta có : $\frac{n+1}{n-2}$ = $\frac{n-2+3}{n-2}$ = 1 + $\frac{3}{n-2}$ 

a, A có giá trị nguyên khi n + 1 chia hết cho n - 2 ⇔ 3 chia hết cho n - 2 

⇒ n - 2 ∈ Ư ( 3 ) = { ± 1 ; ± 3 }

⇒ n ∈ { 1 ; 3 ; - 1 ; 5 }

Bài 2:

\(\text{Vậy tập hợp giá trị của x là:}\)