\(A=\dfrac{x+2}{x+1}=1+\dfrac{1}{x+1}\)

Để A nguyên : 

\(x+1\inƯ\left(1\right)\\ Ư\left(1\right)=\left\{1;-1\right\}\\ \Rightarrow\left\{{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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

a) Để \(\frac{31}{x-1}\)là số nguyên thì 31 chia hết cho x-1

x nguyên => x-1 nguyên => x-1 thuộc Ư (31)={-31;-1;1;31}
Ta có bảng

b)c) Làm tương tự

d) \(\frac{x+3}{x-2}=\frac{x-2+5}{x-2}=1+\frac{5}{x-2}\)

Để \(\frac{x+3}{x-2}\)nguyên thì \(\frac{5}{x-2}\)nguyên

x nguyên => x-2 nguyên

=> x-2 thuộc Ư (5)={-5;-1;1;5}
Ta có bảng

e)f) Phân tích làm tương tự

Bài làm:

c) \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=-\frac{7}{6}\)

\(\Leftrightarrow-\frac{2}{5}+\frac{5}{2}-\frac{4}{9}x=-\frac{7}{6}\)

\(\Leftrightarrow\frac{4}{9}x=-\frac{2}{5}+\frac{5}{2}+\frac{7}{6}\)

\(\Leftrightarrow\frac{4}{9}x=\frac{49}{15}\)

\(\Leftrightarrow x=\frac{49}{15}\div\frac{4}{9}\)

\(\Rightarrow x=\frac{147}{20}\)

Vậy \(x=\frac{147}{20}\)

Bài 2:

a) Ta có: \(F=\frac{3x-2}{x+3}=\frac{\left(3x+9\right)-11}{x+3}=3-\frac{11}{x+3}\)

Để F nguyên \(\Rightarrow\frac{11}{x+3}\inℤ\Leftrightarrow x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)

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

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

2b) Tách

\(G=\frac{x^2-2x+4}{x+1}=\frac{x^2+x-3x-3+7}{x+1}=\frac{x\left(x+1\right)-3\left(x+1\right)+7}{x+1}\)

\(=\frac{x\left(x+1\right)}{x+1}-\frac{3\left(x+1\right)}{x+1}+\frac{7}{x+1}=x-3+\frac{7}{x+1}\)

G là số nguyên <=> \(\frac{7}{x+1}\)là số nguyên <=> \(7⋮x+1\)<=> \(x+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

<=> \(x\in\left\{0;-2;6;-8\right\}\)

\(x^2-2x+\frac{4}{x}+1\)

\(=\left(x^2-2x+1\right)+\frac{4}{x}\)

\(=\left(x-1\right)^2+\frac{4}{x}\)

Để  biểu thức trên là số nguyên thì \(x\inƯ\left(4\right)\)

=> x= {-4;-2;-1;1;2;4}

Giúp mình 

mình ghi hơi nhầm câu ạ là 3x nhé mn. Cảm ơn