a.
ĐKXĐ: \(x\ne\left\{-\dfrac{3}{2};\dfrac{1}{2};\dfrac{5}{2};\dfrac{7}{4};4\right\}\)
\(P=\left(\dfrac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\dfrac{2x-8}{\left(x-4\right)\left(2x-5\right)}-\dfrac{3}{2x-1}\right):\dfrac{-\left(4x-7\right)\left(2x+3\right)}{\left(2x-1\right)\left(2x+3\right)}+1\)
\(=\left(\dfrac{\left(2x-3\right)\left(x-4\right)-\left(2x-8\right)\left(2x-1\right)-3\left(x-4\right)\left(2x-5\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}\right):\dfrac{-\left(4x-7\right)}{2x-1}+1\)
\(=\left(\dfrac{-8x^2+46x-56}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}\right).\dfrac{2x-1}{-\left(4x-7\right)}+1\)
\(=\dfrac{-2\left(x-4\right)\left(4x-7\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\dfrac{\left(2x-1\right)}{-\left(4x-7\right)}+1\)
\(=\dfrac{2}{2x-5}+1=\dfrac{2x-3}{2x-5}\)
b.
\(\left|x\right|=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\Rightarrow P=\dfrac{2}{3}\\x=\dfrac{1}{2}\Rightarrow P=\dfrac{1}{2}\end{matrix}\right.\)
c.
\(P=1+\dfrac{2}{2x-5}\in Z\Rightarrow\dfrac{2}{2x-5}\in Z\)
\(\Rightarrow2x-5\inƯ\left(2\right)\)
Mà 2x-5 luôn lẻ \(\Rightarrow2x-5\in\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{2;3\right\}\)
