a) \(F=\frac{3x-2}{x+3}\)là số nguyên

\(\Leftrightarrow3x-2⋮x+3\)

\(\Leftrightarrow3x+9-11⋮x+3\)

\(\Leftrightarrow3\left(x+3\right)-11⋮x+3\)

\(\Leftrightarrow11⋮x+3\)\(\Leftrightarrow x+3\in\left\{-11;-1;1;11\right\}\)

\(\Leftrightarrow x\in\left\{-14;-4;-2;8\right\}\)

b) \(\frac{x^2-2x+4}{x+1}\)là số nguyên 

\(\Leftrightarrow x^2-2x+4⋮x+1\)

\(\Leftrightarrow x^2+x-3x-3+7⋮x+1\)

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)+7⋮x+1\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)+7⋮x+1\)

\(\Leftrightarrow7⋮x+1\)\(\Leftrightarrow x+1\in\left\{-7;-1;1;7\right\}\)

\(\Leftrightarrow x\in\left\{-8;-2;0;6\right\}\)

\(xy-2x+y=1\)

\(\Leftrightarrow xy-2x+y-2=1-2\)

\(\Leftrightarrow x\left(y-2\right)+y-2=-1\)

\(\Leftrightarrow\left(y-2\right)\left(x+1\right)=-1\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(0;1\right);\left(-2;3\right)\)

a) Ta có A = \(\frac{x-10}{x-5}=\frac{x-5-5}{x-5}=1-\frac{5}{x-5}\)

Vì \(1\inℤ\Rightarrow\frac{-5}{x-5}\inℤ\)

=> \(-5⋮x-5\)

=> x - 5 \(\in\)Ư(-5)

=> \(x-5\in\left\{1;5;-1;-5\right\}\)

=> \(x\in\left\{6;11;4;0\right\}\)

Vậy khi \(x\in\left\{6;11;4;0\right\}\)thì A là số hữu tỉ 

b) Ta có B = \(\frac{3x-2}{x-5}=\frac{3x-15+13}{x-5}=\frac{3\left(x-5\right)+13}{x-5}=3+\frac{13}{x-5}\)

Vì \(3\inℤ\Rightarrow\frac{13}{x-5}\inℤ\)

=> \(13⋮x-5\)

=> \(x-5\inƯ\left(13\right)\Rightarrow x-5\in\left\{1;13;-1;-13\right\}\)

=> \(x\in\left\{6;18;4;-8\right\}\)

Vậy khi  \(x\in\left\{6;18;4;-8\right\}\)thì B là số hữu tỉ

c) Ta có C = \(\frac{x-3}{2x}\)

=> 2C = \(\frac{2x-6}{2x}=1-\frac{6}{2x}=1-\frac{3}{x}\)

Vì \(1\inℤ\Rightarrow\frac{3}{x}\inℤ\Rightarrow3⋮x\Rightarrow x\inƯ\left(3\right)\Rightarrow x\in\left\{1;3;-1;-3\right\}\)

Vậy khi \(x\in\left\{1;3;-1;-3\right\}\)thì  C là số hữu tỉ

Ta có:

\(T=\frac{3x-8}{x-5}=\frac{3x-15+7}{x-5}=\frac{3.\left(x-5\right)+7}{x-5}=\frac{3.\left(x-5\right)}{x-5}+\frac{7}{x-5}=3+\frac{7}{x-5}\)

Để T nguyên thì \(\frac{7}{x-5}\) nguyên

\(\Rightarrow x-5\inƯ\left(7\right)\)

\(\Rightarrow x-5\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{6;4;12;-2\right\}\)

Vậy \(x\in\left\{6;4;12;-2\right\}\) thì T nguyên

a, \(\dfrac{x}{7}\) \(\in\) Q     ⇔ \(x\in z\)

b, \(\dfrac{5}{x}\) \(\in\) Q     ⇔ \(x\) \(\ne\) 0; \(x\) \(\in\)  Z

c, - \(\dfrac{5}{2x}\)  \(\in\) Q ⇔ \(x\) \(\ne\) 0; \(x\in Z\)

a) Tập hợp số nguyên chia hết cho 7 là

\(\Rightarrow x\in A=\left\{\pm7;\pm14;\pm21;...\right\}\)

\(\Rightarrow A=\left\{x\inℕ|x=\pm7k;k\inℤ\right\}\)

Vậy để \(\dfrac{x}{7}\in Q\)

\(\Rightarrow x\in A\)

b) \(\dfrac{5}{x}\inℚ\)

\(\Rightarrow x\in\left\{\pm1;\pm5\right\}\)

c) \(-\dfrac{5}{2x}\inℚ\)

\(\Rightarrow2x\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow x\in\left\{\pm\dfrac{1}{2};\pm\dfrac{5}{2}\right\}\)

\(\Rightarrow x\in\varnothing\)

Bài 1:

a) \(x=\frac{a+1}{a+9}=\frac{a+9-8}{a+9}=\frac{a+9}{a+9}-\frac{8}{a+9}=1-\frac{8}{a+9}\)

Để \(x\in Z\)thì \(a+9\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Vậy \(a\in\left\{-17;-13;-11;-10;-8;-7;-5;-1\right\}\)

b) \(x=\frac{a-1}{a+4}=\frac{a+4-5}{a+4}=\frac{a+4}{a+4}-\frac{5}{a+4}=1-\frac{5}{a+4}\)

Để \(x\in Z\)thì \(a+4\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Vậy \(a\in\left\{-9;-5;-3;1\right\}\)

Bài 2:

a) \(t=\frac{3x-8}{x-5}=\frac{3x-15}{x-5}+\frac{7}{x-5}=\frac{3\left(x-5\right)}{x-5}+\frac{7}{x-5}=3+\frac{7}{x-5}\)

Để \(t\in Z\)thì \(x-5\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Vậy \(x\in\left\{-2;4;6;12\right\}\)

b)\(q=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=\frac{2\left(x-3\right)}{x-3}+\frac{7}{\left(x-3\right)}=2+\frac{7}{x-3}\)

Để \(q\in Z\)thì \(x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Vậy \(x\in\left\{-4;2;4;10\right\}\)

c)\(p=\frac{3x-2}{x+3}=\frac{3x+9}{x+3}-\frac{11}{x+3}=\frac{3\left(x+3\right)}{x+3}-\frac{11}{x+3}=3-\frac{11}{x+3}\)

Để \(p\in Z\)thì \(x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)

Vậy \(x\in\left\{-14;-4;-2;8\right\}\)

Bài 3:

Gọi \(d\inƯC\left(2m+9;14m+62\right)\)

\(\Rightarrow\hept{\begin{cases}\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}7\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(14m+63\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)

\(\Rightarrow\left[\left(14m+63\right)-\left(14m+62\right)\right]⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

\(\RightarrowƯC\left(2m+9;14m+62\right)=1\)

Vậy \(x=\frac{2m+9}{14m+62}\)là p/s tối giản

Để \(\frac{-8}{x+8}\)là số hữu tỉ dương

\(\Leftrightarrow x+8< 0\)

\(\Leftrightarrow x< -8\)

Vậy x < -8