\(ĐKXĐ:x\ne\pm1\)

a) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)

\(\Leftrightarrow P=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+3x+5}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow P=\frac{2x^2+x-3-x^2-3x-2+3x+5}{\left(x-1\right)\left(x+1\right)}\)

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

\(\Leftrightarrow P=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow P=\frac{x}{x-1}\)

b) Để \(P\inℤ\)

\(\Leftrightarrow x⋮x-1\)

\(\Leftrightarrow x-1+1⋮x-1\)

\(\Leftrightarrow1⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{0;2\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)

a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)

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

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6

ĐKXĐ \(x\ne0;x\ne1;x\ne-1\)

\(A=\frac{\left(x+1+1-x\right)}{\left(1-x^2\right)-\frac{5-x}{1-x^2}}:\frac{\left(1-2x\right)}{x^2-1}\)

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

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

\(A=\frac{\left(3x-2\right)}{1-2x}\)

\(a,ĐKXĐ:x\ne\pm1;x\ne\frac{1}{2}\)

\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^{^2}}\right):\frac{1-2x}{x^2-1}\)

\(=\left(\frac{1}{x-1}+\frac{2}{x+1}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x+1+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}:\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)

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

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

\(b,Vớix\ne\pm1;x\ne\frac{1}{2}\)ta có \(A=\frac{2x+4}{1-2x}=\frac{-1\left(1-2x\right)+5}{1-2x}=-1+\frac{5}{1-2x}\)

Với x thuộc Z để A nguyên thì \(5⋮1-2x\Rightarrow1-2x\inƯ\left\{5\right\}=\left\{\pm1;\pm5\right\}\)

Với 1-2x=1 => x= 0(TMĐKXĐ)

với 1-2x=-1 => x=1(loại)

với 1-2x=5 => x=-2(tmđkxđ)

với 1-2x=-5 => x=3(tmđkxđ)

Vậy với \(x\in\left\{0;-2;-3\right\}\)thì A nguyên

a) A xác định \(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x+1\ne0\\1-2x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\2x\ne1\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne\frac{1}{2}\end{cases}\Leftrightarrow}x\ne\pm1;x\ne\frac{1}{2}}\)

\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)

\(A=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(A=\frac{x+1+2x-1+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(A=\frac{\left(2x+5\right)\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}=\frac{2x+5}{1-2x}\)

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

<=> M =