a) \(A=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right):\frac{x^2-6x+9}{\left(2-x\right)\left(x-3\right)}\)(ĐKXĐ: \(\hept{\begin{cases}x\ne\pm2\\x\ne3\end{cases}}\))\(=\left[\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{4-x^2}\right]:\frac{\left(x-3\right)^2}{\left(2-x\right)\left(x-3\right)}\)\(=\frac{4x\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}.\frac{2-x}{x-3}=\frac{4x}{x-3}\)

b) l\(x-5\)l\(=2\Leftrightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\left(n\right)\\x=3\left(l\right)\end{cases}\Rightarrow A=\frac{4.7}{7-3}=\frac{28}{4}=7}\)
c)
* Để A có giá trị là một số nguyên thì \(A=\frac{4x}{x-3}=\frac{4x-12+12}{x-3}=4+\frac{12}{x-3}\)là một số nguyên hay \(\frac{12}{x-3}\)là một số nguyên \(\Rightarrow x-3\inƯ\left(12\right)\Rightarrow S=\left(-9;-3;-1;0;1;4;5;6;7;9;15\right)\)(1)
* Để \(A=4+\frac{12}{x-3}< 4\Leftrightarrow\frac{12}{x-3}< 0\) thì \(x-3< 0\Leftrightarrow x< 3\)(2)
(1)(2) \(\Rightarrow S=\left(-9;-3;-1;0;1\right)\)

a) 

Thay x = -1 ( thỏa mãn ĐKXĐ ) vào biểu thức B , ta có :

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

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

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

\(A=\frac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\)

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

c) Ta có : 

\(P=A.B\)

\(P=\frac{3x}{\left(x-2\right)\left(x+2\right)}.\frac{2-x}{x}\)

Mà P = 1/2

\(\Leftrightarrow\frac{3x}{\left(x-2\right)\left(x+2\right)}.\frac{-\left(x-2\right)}{x}=\frac{1}{2}\)

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

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

\(\Leftrightarrow x+2=-6\Leftrightarrow x=-8\)( thỏa mãn )

d) P nguyên dương

\(\Leftrightarrow\frac{-3}{x+2}\)nguyên dương

<=> x + 2 thuộc Ư(3) { -1 ; -3 }

Bảng tìm x

Vậy ....................

cảm ơn bn nhé nhg mk hỏi sao x +2x+ x= 3x đc z mk tưởng là 4x

a) \(-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{15}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le\frac{-33}{15}:\frac{21}{15}\)

=> \(-10\le x\le\frac{-11}{7}\)

=> \(x\in\left\{-10;-9,-8,-7,-6,-5,-4,-3,-2,-1\right\}\)

-4/1/3.1/3< x <  -2/3.-11/12

-1/4/9< x < 11/18

-26/18< x < 11/18

Vậy x={-26/18;-25/18;.............;11/18}

Sao đề khó đọc thế. Bạn viết rõ lại đi

Số 0 nha bn !

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{13}{12}\le x\le\frac{1}{24}-\left(-\frac{5}{24}\right)\)

\(\Rightarrow\frac{6}{12}-\frac{13}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow x\in\left\{-7;-6;-5;...;0;1;2;3\right\}\)

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\le x\le\frac{1}{24}-\frac{1}{8}+\frac{1}{3}\)

\(\Rightarrow\frac{6-4-9}{12}\le x\le\frac{1-3+8}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{6}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow-7\le x\le3\)

\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3;....;2;3\right\}\)