\(P=\left(\frac{2+x}{x^2+2x+1}-\frac{x-2}{x^2-1}\right).\frac{1-x^2}{x}\)
a) ĐKXĐ:
\(\hept{\begin{cases}x^2+2x+1\ne0\\x^2-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\ne0\\\left(x-1\right)\left(x+1\right)\ne0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x\ne1;x\ne-1\\x\ne0\end{cases}}}\)
<=> x khác -1
x khác 1; x khác -1
x khác 0
<=> x khác -1;1;0
Vậy ĐKXĐ là x khác -1;1;0
b) \(P=\left(\frac{2+x}{x^2+2x+1}-\frac{x-2}{x^2-1}\right).\frac{1-x^2}{x}\)
\(\Rightarrow P=\left(\frac{2+x}{\left(x+1\right)^2_{x-1}}-\frac{x-2}{\left(x-1\right)\left(x+1\right)_{x+1}}\right).\frac{1-x}{x}\)
MTC: (x+1)^2(x-1)
\(\Rightarrow P=\left(\frac{\left(2+x\right)\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}-\frac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\right).\frac{1-x}{x}\)
\(\Rightarrow P=\left(\frac{2x-2+x^2-x}{\left(x+1\right)^2\left(x-1\right)}-\frac{x^2+x-2x-2}{\left(x+1\right)^2\left(x-1\right)}\right)\frac{1-x}{x}\)
\(\Rightarrow P=\left(\frac{x-2+x^2-x^2+x+2}{\left(x+1\right)^2\left(x-1\right)}\right).\frac{1-x}{x}\)
\(\Rightarrow P=\frac{2x}{\left(x+1\right)^2\left(x-1\right)}.\frac{1-x}{x}\)
\(\Rightarrow P=\frac{2x}{-\left(1-x\right)\left(x+1\right)^2}.\frac{1-x}{x}\)
\(\Rightarrow P=-\frac{x}{\left(x+1\right)^2}\) (tmđkxđ)
c)
\(P=-\frac{x}{\left(x+1\right)^2}=-\frac{x+1-1}{\left(x+1\right)\left(x+1\right)}=-\frac{x+1}{x+1}-\frac{1}{x+1}=-1-\frac{1}{x+1}\) ( ĐKXĐ là x khác -1;1;0) \(\left(P\in Z\right)\)
\(P\in Z\Leftrightarrow\frac{-1}{x+1}\)
Nên x+1 thuộc Ư(-1)={1;-1)
x+1=1=>x=1-1=0 ( o t/m đk)
x+1=-1=>x=-1-1=-2( (t/m đk)
<=> x thuộc -2 thì gt của BT P là số nguyên
