a, ĐK : \(x\ne\pm3;\frac{1}{2}\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)

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

\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)

\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)

b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)

TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)

Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)

TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)

Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)

c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)

\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)

\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

a)     \(ĐKXĐ:x\ne-3;x\ne2\)

b)     \(P=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(P=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

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

\(P=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

vậy \(P=\frac{x-4}{x-2}\)

\(P=\frac{-3}{4}\) \(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3.\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\)

c) \(P\in Z\Leftrightarrow\frac{x-4}{x-2}\in Z\)

\(\frac{x-2-6}{x-2}=1-\frac{6}{x-2}\in Z\)

mà \(1\in Z\Rightarrow\left(x-2\right)\inƯ\left(6\right)\in\left(\pm1;\pm2;\pm3;\pm6\right)\)

mà theo ĐKXĐ:  \(\Rightarrow\in\left(\pm1;-2;3;\pm6\right)\)

thay mấy cái kia vào rồi tìm \(x\)

d) \(x^2-9=0\Rightarrow x^2=9\Rightarrow x=\pm3\)

khi \(x=3\Rightarrow P=\frac{3-4}{3-2}=-1\)

khi \(x=-3\Rightarrow P=\frac{-3-4}{-3-2}=\frac{-7}{-5}=\frac{7}{5}\)

a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)

Kết hợp với giả thiết vậy x = -1 

Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)

c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )

d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)

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

e, tự làm nhéee 

f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)

mà x + 4 > x + 2 

\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)

Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )

Vậy ko có x nguyên để C < 0 

g, Ta có :  \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)

Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)

\(=\left(x+1\right)^2-9\ge-9\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTNN D là -9 khi x = -1 

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

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

\(\Leftrightarrow x< 2\)

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

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

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

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

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

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

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