a, đk : x khác 5 ; -5
\(P=\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{10x-25}{x^2+5x}+\dfrac{x}{5-x}\)
\(=\left(\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right):\dfrac{5\left(2x-5\right)}{x\left(x+5\right)}+\dfrac{x}{5-x}\)
\(=\left(\dfrac{10x-25}{x\left(x-5\right)\left(x+5\right)}\right):\dfrac{5\left(2x-5\right)}{x\left(x+5\right)}+\dfrac{x}{5-x}\)
\(=\dfrac{1}{x-5}-\dfrac{x}{x-5}=\dfrac{1-x}{x-5}\)
c, Ta có : \(P=\dfrac{1-x}{x-5}=2013\Rightarrow1-x=2013x-10065\Leftrightarrow2014x=10066\Leftrightarrow x=4,99\)
d, \(\dfrac{1-x}{x-5}=\dfrac{-\left(x-1\right)}{x-5}=\dfrac{-\left(x-5+4\right)}{x-5}=\dfrac{-\left(x-5\right)+4}{x-5}=-1+\dfrac{4}{x-5}\)
\(\Rightarrow x-5\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| x - 5 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | 6 | 4 | 7 | 3 | 9 | 1 |
a, ĐKXĐ:\(\left\{{}\begin{matrix}x^2-25\ne0\\x^2+5x\ne0\\10x-25\ne0\\5-x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\\x\ne\dfrac{5}{2}\end{matrix}\right.\)
b, \(P=\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{10x-25}{x^2+5x}+\dfrac{x}{5-x}\)
\(\Rightarrow P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\)
\(\Rightarrow P=\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}:\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\)
\(\Rightarrow P=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}:\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\)
\(\Rightarrow P=\dfrac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\dfrac{x\left(x+5\right)}{10x-25}-\dfrac{x}{x-5}\)
\(\Rightarrow P=\dfrac{1}{x-5}-\dfrac{x}{x-5}\)
\(\Rightarrow P=\dfrac{1-x}{x-5}\)
c,\(P=2013\)
\(\Leftrightarrow\dfrac{1-x}{x-5}=2013\)
\(\Leftrightarrow2013\left(x-5\right)=1-x\\ \Leftrightarrow2013x-10065-1+x=0\\ \Leftrightarrow2014x-10066=0\\ \Leftrightarrow x=\dfrac{5033}{1007}\)
d, \(P\in Z\)
\(\Leftrightarrow\dfrac{1-x}{x-5}\in Z\)
\(\Leftrightarrow\dfrac{-4-x+5}{x-5}\in Z\)
\(\Leftrightarrow\dfrac{-4-\left(x-5\right)}{x-5}\in Z\)
\(\Leftrightarrow\dfrac{4}{5-x}-1\in Z\)
\(Vì.-1\in Z\Rightarrow5-x\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\Rightarrow x\in\left\{9;7;6;4;3;1\right\}\)
