\(a,P=\left(\dfrac{1}{x-2}-\dfrac{4}{x^2-4}\right):\dfrac{x+1}{2x+4}\\ =\left(\dfrac{1.\left(x+2\right)-4}{\left(x-2\right)\left(x+2\right)}\right)\times\dfrac{2\left(x+2\right)}{x+1}\\ =\dfrac{x+2-4}{\left(x-2\right)\left(x+2\right)}\times\dfrac{2\left(x+2\right)}{x+1}\\ =\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\times\dfrac{2\left(x+2\right)}{x+1}\\ \\ =\dfrac{2}{x+1}\)
\(b,\) Để \(P\) nguyên
\(\Rightarrow x+1\inƯ\left(2\right)\\ Ư\left(2\right)=\left\{1;-1;2;-2\right\}\\ \Rightarrow\left\{{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\left(t/m\right)\\x=-2\left(kot/m\right)\\x=1\left(t/m\right)\\x=-3\left(t/m\right)\end{matrix}\right.\)
Vậy \(x=\left\{0;1;3\right\}\)
\(a,P=\left(\dfrac{1}{x-2}-\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{2\left(x+2\right)}{x+1}\)
\(=\dfrac{x+2-4}{\left(x-2\right)\left(x+2\right)}.\dfrac{2\left(x+2\right)}{x+1}\)
\(=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2}{x+1}\)
\(b,\) Để P có giá trị nguyên thì \(2⋮\left(x+1\right)\) \(\Leftrightarrow x+1\inƯ\left(2\right)\left\{\pm1;\pm2\right\}\)
\(x+1=1\Leftrightarrow x=0\left(n\right)\)
\(x+1=-1\Leftrightarrow x=-2\left(l\right)\)
\(x+1=2\Leftrightarrow x=1\left(n\right)\)
\(x+1=-2\Leftrightarrow x=-3\left(n\right)\)
Kết các TH \(\Rightarrow x=\left\{0;-3;1\right\}\)
a) \(P=\left(\dfrac{1}{x-2}-\dfrac{4}{x^2-4}\right):\dfrac{x+1}{2x+4}\)
\(P=\left(\dfrac{1}{x-2}-\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{x+1}{2\left(x+2\right)}\)
\(P=\left[\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right].\dfrac{2\left(x+2\right)}{x+1}\)
\(P=\dfrac{1}{x+2}.\dfrac{2\left(x+2\right)}{x+1}\)
\(P=\dfrac{2}{x+1}\)
b) Để P có giá trị nguyên thì \(x+1\inƯ\left(2\right)\)
Mà \(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x+1 | -2 | -1 | 1 | 2 |
| x | -3 | -2 | 0 | 1 |
| nhận | loại | nhận | nhận |
Vậy khi \(x\in\left\{-3;0;1\right\}\) để P có giá trị nguyên
