\(\frac{3a+45}{a+9}\)là số nguyên

\(\Rightarrow3a+45⋮a+9\)

Ta có : \(3a+45⋮a+9\)

\(\Rightarrow3a+27+18⋮a+9\)

\(\Rightarrow3\left(a+9\right)+18⋮a+9\)

\(\Rightarrow18⋮a+9\)

\(\Rightarrow a+9\inƯ\left(18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\)

\(\Rightarrow a\in\left\{-27;-18;-15;-12;-11;-10;-8;-7;-6;-3;0;9\right\}\)

Học tốt!

Để phân số trên là số nguyên thì -19 phải chia hết cho a+8

=>a+8\(\in\)Ư(-19)

=>a+8\(\in\){1; -1; 19; -19}

KL:a\(\in\){-7; -9; 11; -27}

để \(\frac{-19}{a+8}\)là số nguyên thì:

a+8\(\in\)Ư(-19)={-1;1;-19;19}

với a+8=-1

a=-9

với a+8=1

a=-7

với a+8=19

a=11

với a+8=-19

a=-27

vậy a={-9;-7;11;-27} thì \(\frac{-19}{a+8}\)là số nguyên

\(A=\frac{7a-2}{a-3}=\frac{7\left(a-3\right)+19}{a-3}=7+\frac{19}{a-3}\)

Để A nguyên thì \(\frac{19}{a-3}\) nguyên

Khi \(a-3\in\left\{1;19;-1;-19\right\}\)

\(\Leftrightarrow a\in\left\{4;22;2;-16\right\}\)

Vậy

Bài 1b:

\(\frac31\) + \(\frac33\) + \(\frac36\) + \(\frac{3}{10}\) + ...+\(\frac{3}{x\left(x+1\right):2}\) = \(\frac{2015}{336}\)

3.(\(\frac11+\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{x\left(x+1):2\right.})\) = \(\frac{2015}{336}\)

3.2(\(\frac12+\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)})=\) \(\frac{2015}{336}\)

6.(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\cdots+\frac{1}{x.\left(x+1\right)})\) = \(\frac{2015}{336}\)

6.(\(\frac11-\frac12\) + \(\frac12\)-\(\frac14\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

6.(\(\frac11\) - \(\frac{1}{x+1}\)) = \(\frac{2015}{336}\)

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{336}\) : 6

1 - \(\frac{1}{x+1}\) = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = 1 - \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\) = \(\frac{1}{2016}\)

\(x+1\) = 2016

\(x\) = 2016 - 1

\(x\) = 2015

Bài 2:

A = \(\frac{6n+1}{4n+3}\) (n ∈ Z\(^{-}\))

A ∈ Z khi và chỉ khi:

(6n + 1) ⋮ (4n + 3)

(12n + 2) ⋮ (4n + 3)

[3(4n + 3) - 7] ⋮ (4n + 3)

7 ⋮ (4n + 3)

(4n + 3) ∈ Ư(7) = {-7; -1; 1; 7}

n ∈ {- 5/2; -1; - 1/2; 1}

Nếu n = - 1 thì A = (-6 + 1)/(-4 + 3) = 5 (loại)

Nếu n = 1 thì: A = (6 + 1).(4+3) = 1 (loại)

Không có giá trị nào thỏa mãn đề bài hay n ∈ ∅

{9;7;10;6;12;4;16;0}

Bài 1 :

x < 0 \(\Leftrightarrow\) 3a - 5 < -2 \(\Leftrightarrow\) 3a < 3 \(\Leftrightarrow\) a < 1

Bài 2 :

a) \(\frac{3a-5}{a}=3+\frac{5}{a}\in Z\)\(\Leftrightarrow a\inƯ\left(5\right)\)

\(\Leftrightarrow a\in\left\{-5;-1;1;5\right\}\)

b) \(\frac{2b-7}{b+2}=\frac{2b+4-11}{b+2}=2-\frac{11}{b+2}\in Z\) \(\Leftrightarrow b+2\inƯ\left(11\right)\)

\(\Leftrightarrow b+2\in\left\{-11;-1;1;11\right\}\)

\(\Leftrightarrow b\in\left\{-13;-3;-1;9\right\}\)

\(\text{Ta có:}\)

\(\text{Để}\)\(\frac{4b+42}{b+7}\)\(\text{nguyên thì}\)\(4b+42⋮b+7\)

\(\text{Lại có:}\)

\(\text{4b + 42 = 4b + 28 + 14 = 4( b+7 ) + 14}\)

\(\text{Vì}\)\(b+7⋮b+7\)\(\Rightarrow4\left(b+7\right)⋮b+7\)

\(\text{Do đó:}\)\(14⋮b+7\)

\(\Rightarrow b+7\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)

\(\Rightarrow b\in\left\{-6;-5;0;7\right\}\)

huhu dap an la no biet

\(A=\frac{5n-19}{n-4}=\frac{5n-20+1}{n-4}=\frac{5\left(n-4\right)+1}{n-4}=5+\frac{1}{n-4}\)

Vì \(5\inℤ\)\(\Rightarrow\)Để \(A\inℤ\)thì \(\frac{1}{n-4}\inℤ\)

\(\Rightarrow1⋮n-4\)\(\Rightarrow n-4\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{3;5\right\}\)

Vậy \(n\in\left\{3;5\right\}\)