a, ĐKXĐ: \(x\ne\pm3\)

\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)

b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)

c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)

\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)

Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)

a, ĐỂ A xác định : 

\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)

\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)

\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)

\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{x-4}{x+3}\)

b

a) \(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{x^2-9}\right):\frac{3}{x-3}\)

\(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

A xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\end{cases}}}\)

b) \(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x^2-3x+2x^2+6x-3x^2-12}{\left(x+3\right)\left(x-3\right)}+\right]:\frac{3}{x-3}\)

\(A=\left[\frac{3x-12}{\left(x+3\right)\left(x-3\right)}\right].\frac{x-3}{3}\)

\(A=\left[\frac{3\left(x-4\right)}{\left(x+3\right)\left(x-3\right)}\right].\frac{x-3}{3}\)

\(A=\frac{x-4}{x+3}\)

Với \(x=-4\)

\(\Rightarrow A=\frac{-4-4}{-4+3}=-\frac{8}{-1}=8\)

Vậy \(A=8\)tại \(x=-4\)

c) \(A=\frac{x-4}{x+3}=\frac{x+3-7}{x+3}=1-\frac{7}{x+3}\)

Có \(1\in Z\)

Để \(A\in Z\Rightarrow\frac{7}{x+3}\in Z\)

Có: \(x\in Z\Rightarrow x+3\in Z\Rightarrow\frac{7}{x+3}\in Z\Leftrightarrow\left(x+3\right)\in\text{Ư}\left(7\right)=\left\{\pm1;\pm7\right\}\)

b tự lập bảng nhé~

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

ỳtct7ct7c7c7t79tc9

\(A=\frac{x^2-10x+36}{x-5}=\frac{x^2-10x+25+9}{x-5}\) \(=\frac{\left(x-5\right)^2+9}{x-5}=x-5+\frac{9}{x-5}\)

để \(A\in Z\)

<=> \(\frac{9}{x-5}\in Z\)mà \(x\in Z\)

=> \(x-5\inƯ\left(9\right)\)

=> \(x-5\in\left(1;-1;3;-3;9;-9\right)\)

=> \(x\in\left(6;4;8;2;14;-4\right)\)

học tốt

a) Để biểu thức nguyên 

\(\Leftrightarrow2x+3⋮x-1\)

\(\Leftrightarrow2.\left(x-1\right)+5⋮x-1\)

Mà \(2.\left(x-1\right)⋮x-1\)

\(\Rightarrow5⋮x-1\)

Tự tìm x

cảm ơn bạn

b) Để  biểu thức nguyên

\(\Leftrightarrow3x+2⋮2x+1\)

\(\Leftrightarrow2.\left(3x+2\right)⋮2x+1\)

\(\Leftrightarrow6x+4⋮2x+1\)

\(\Leftrightarrow3.\left(2x+1\right)+1⋮2x+1\)

Mà \(3.\left(2x+1\right)⋮2x+1\)

\(\Rightarrow1⋮2x+1\)

Tìm nốt x

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).