Lời giải:
\(2n+5\vdots n-3\)
\(\Leftrightarrow 2(n-3)+11\vdots n-3\)
\(\Leftrightarrow 11\vdots n-3\Rightarrow n-3\in \text{Ư}(11)\)
\(\Rightarrow n-3\in\left\{\pm 1; \pm 11\right\}\Rightarrow n\in \left\{-8; 2; 4; 14\right\}\)
Vậy.........
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\(5n-7\vdots 2n\)
\(\Rightarrow 2(5n-7)\vdots 2n\)
\(\Leftrightarrow 5.2n-14\vdots 2n\)
\(\Leftrightarrow 14\vdots 2n\Rightarrow 7\vdots n\Rightarrow n\in \text{Ư}(7)\)
\(\Rightarrow n\in\left\{\pm 1; \pm 7\right\}\)
Thử lại thấy đều thỏa mãn. Vậy..............
2n+5⋮n−3
2n+5⋮n−3
⇔2(n−3)+11⋮n−3⇔2(n−3)+11⋮n−3
⇔11⋮n−3⇒n−3∈Ư(11)⇔11⋮n−3⇒n−3∈Ư(11)
⇒n−3∈{±1;±11}⇒n∈{−8;2;4;14}⇒n−3∈{±1;±11}⇒n∈{−8;2;4;14}
Vậy.........
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5n−7⋮2n5n−7⋮2n
⇒2(5n−7)⋮2n⇒2(5n−7)⋮2n
⇔5.2n−14⋮2n⇔5.2n−14⋮2n
⇔14⋮2n⇒7⋮n⇒n∈Ư(7)⇔14⋮2n⇒7⋮n⇒n∈Ư(7)
⇒n∈{±1;±7}
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