d, A nguyên dương <=> \(\frac{-4}{2+x}\) nguyên dương

                                <=> \(\begin{cases}2+x< 0\\2+x\inƯ4\end{cases}\)

                                 <=> 2 + x \(\in\) {-1; -2; -4}

Thay 2 + x = -1 => x = -3

2 + x = -2 => x = -4

2 + x = -4 => x = -6

Vây x \(\in\left\{-3;-4;-6\right\}\) 

a) x khác 0 ; 2 ;-2

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

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

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

b) Ta có: 2x²+x=0

<=>x.(2x+1)=0

<=>x=0 (loại) hoặc x=-1/2

Khi x=-1/2 => A=\(-\frac{4}{-\frac{1}{2}+2}=-\frac{8}{3}\)

c)Để A=1/4

Thì: \(-\frac{4}{x+2}=\frac{1}{4}\Rightarrow x+2=-16\Leftrightarrow x=-18\)(nhận)

Vậy x=-18 thì A=1/4

d)Để A nguyên dương thì x+2 thuộc ước âm của 4

=>x+2=-1 hoặc x+2=-2 ; hoặc x+2=-4

=>x=-3 hoặc x=-4 hoặc x=-6

Vậy x=-3 hoặc x=-4 hoặc x=-6 thì A nguyên dương

a)điều kiện xác định x\(\ne\left\{\pm2,0\right\}\)

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

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

=\(\frac{-4}{2+x}\)

b) 2x²+x=0<=> x=0 hoặc x=-2

ta thấy x=o và x=-2 không thảo điều kện => k có giá trị A nào thỏa mãn

c)A=1/4<=>\(\frac{-4}{2+x}\)=1/4<=> -16=2+x<=> x= - 18

d)  \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)

\(\Leftrightarrow x-2< 0\)  ( vì \(-1< 0\))

\(\Leftrightarrow x< 2\)

\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

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

  \(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

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

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

theo câu a) \(A=\frac{-1}{x-2}\)  với ĐKXĐ: \(x\ne\pm2\)

b) \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)   \(\Rightarrow x=-1\)  ( vì \(x=2\)  ko TM ĐKXĐ )

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

vậy khi \(x=-1\)  thì \(A=\frac{1}{3}\)

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

đkxd: \(x\ne\left\{\pm3\right\}\)

a) B= \(\frac{21+\left(x-4\right)\left(x+3\right)-\left(x+1\right)\left(x-3\right)}{x^2-9}:\left(\frac{x+3-1}{x+3}\right)\)

=\(\frac{21+x^2-x-12-x^2+2x+3}{x^2-9}.\frac{x+3}{x+2}\)

=\(\frac{x+12}{x-3}\)

b)|2x+1|=5

<=> \(\left[\begin{array}{nghiempt}2x+1=-5\\2x+1=5\end{array}\right.\)<=> x=-3 hoặc x=2

với x=-3 thì B=\(\frac{-3}{2}\)

với x=2 thì B=-14