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a) Để \(\frac{7}{n+1}\) đạt giá trị nguyên
<=> 7 \(⋮\) ( n + 1 )
=> n + 1 \(\in\) Ư(7) = { - 7 ; -1 ; 1 ; 7 }
=> n \(\in\) { -8 ; -2 ; 0 ; 6 }
b) Để \(\frac{n+5}{n-2}\) đạt giá trị nguyên
<=> \(n+5⋮n-2\)
=> ( n - 2 ) + 7 \(⋮\) n - 2
=> 7 \(⋮\) n - 2
=> n - 2 \(\in\) Ư(7) = { - 7 ; - 1 ; 1 ; 7 }
=> n \(\in\) { - 5 ; 1 ; 3 ; 9 }
c) Để \(\frac{4n-1}{n-3}\) đạt giá trị nguyên
<=> 4n-1 \(⋮\) n - 3
=> ( 4n - 12 ) + 11 \(⋮\) n- 3
=> 4(n-3) + 11 \(⋮\) n - 3
=> 11 \(⋮\)n - 3
=> n - 3 \(\in\) Ư(11) = { - 11 ; - 1 ; 1 ; 11}
=> n \(\in\) { - 8 ; 2 ; 4 ; 14 }
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360.
Ta có:
S = \(\dfrac{a+b}{c}\)+\(\dfrac{b+c}{a}\)+\(\dfrac{c+a}{b}\)
S = \(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\)
S = \(\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
Vì \(\dfrac{a}{c}+\dfrac{c}{a}\ge2\); \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\); \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow\)S \(\ge\) 2 + 2 + 2
\(\Rightarrow\)S \(\ge\) 6
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