1. \(\frac{x+2}{5}=\frac{3x-2}{2}\)

=> 2(x + 2) = 5(3x - 2)

=> 2x + 4 = 15x - 10

=> 2x - 15x = -10 - 4

=> -13x = -14

=> x = 13/4

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

<=> 2x+4=15x-10

<=> 2x-15x=-10-4

<=> -13x=-14

<=> x=\(\frac{14}{13}\)

Bài 2: xy+2x+y=0

<=> (xy+2x)+(y+2)=2

<=> x(y+2)+(y+2)=2

<=> (y+2)(x+1)=2

Vì x,y nguyên => y+2; x+1 nguyên => y+2; x+1 nguyên 

=> y+2; x+1 \(\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

ta có bảng

1,b, 2xy - x = y + 5

<=> 4xy - 2x = 2y + 10

<=> 2x(2y - 1) - (2y - 1) = 11

<=> (2x - 1)(2y - 1) = 11

Lập bảng ra làm nốt

\(1,c,\frac{1}{x}-3=-\frac{1}{y-2}\)

\(\Leftrightarrow y-2-3x\left(y-2\right)=-x\)

\(\Leftrightarrow y-2-3xy+6x+x=0\)

\(\Leftrightarrow-3xy+7x+y-2=0\)

\(\Leftrightarrow-x\left(3y-7\right)+y-2=0\)

\(\Leftrightarrow-3x\left(3y-7\right)+3y-6=0\)

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

\(\Leftrightarrow\left(1-3x\right)\left(3y-7\right)=-1\)

Lập bảng làm nốt

a.

để x \(\in Z\) thì \(2x-3\inƯ_{\left(7\right)}\left\{1;-1;7;-7\right\}\)

Vậy x ={2;1;5;-2}

câu b mik ko pt lm đâu nhé sorry

c.

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

để x \(\in Z\) thì \(x+1\inƯ_{\left(1\right)}\left\{1;-1\right\}\)

câu d giống câu a rồi nhé

a, Với x = 1 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot1+2}{1-3}=\frac{5}{-2}=\frac{-5}{2}\)

Với x = 2 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot2+2}{2-3}=\frac{8}{-1}=-\frac{8}{1}=-8\)

Với x =\(\frac{5}{2}\)thì : \(A=\frac{3x+2}{x-3}=\frac{3\cdot\frac{5}{2}+2}{\frac{5}{2}-3}=\frac{\frac{15}{2}+2}{\frac{5}{2}-3}=\frac{\frac{19}{2}}{-\frac{1}{2}}=\frac{19}{2}\cdot(-2)=\frac{19}{1}\cdot(-1)=-19\)

b, Ta có : \(\frac{3x+2}{x-3}=\frac{3x-9+11}{x-3}=\frac{3(x-3)+11}{x-3}=3+\frac{11}{x-3}\)

\(\Leftrightarrow11⋮x-3\Leftrightarrow x-3\inƯ(11)=\left\{\pm1;\pm11\right\}\)

Lập bảng :

c,Để suy nghĩ đã

Làm tiếp :v

c, \(B=\frac{x^2+3x-7}{x+3}=\frac{x(x+3)-7}{x+3}=x-\frac{7}{x+3}\)

\(\Rightarrow7⋮x+3\Leftrightarrow x+3\inƯ(7)=\left\{\pm1;\pm7\right\}\)

Lập bảng :

d, Tương tự

a) Ta có :

\(2x-3⋮x-1\)

Mà \(x-1⋮x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3⋮x-1\\2x-2⋮x-1\end{matrix}\right.\)

\(\Leftrightarrow1⋮x-1\)

Vì \(x\in Z\Leftrightarrow x-1\in Z;x-1\inƯ\left(1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\Rightarrow x=0\left(tm\right)\\x-1=-1\Leftrightarrow x=-2\left(tm\right)\end{matrix}\right.\)

Vậy ...............

b) Ta có :

\(3x+2⋮2x-1\)

Mà \(2x-1⋮2x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+4⋮2x-1\\6x-3⋮2x-1\end{matrix}\right.\)

\(\Leftrightarrow7⋮2x-1\)

Vì \(x\in Z\Leftrightarrow2x-1\in Z;2x-1\inƯ\left(7\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=7\Leftrightarrow x=4\left(tm\right)\\2x-1=-7\Leftrightarrow x=-4\left(tm\right)\end{matrix}\right.\)

Vậy ............