Câu a bị lỗi đề , sửa Cm \(P=\dfrac{x}{x+2}\)
\(a,=\left(\dfrac{3}{2\left(x+2\right)}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right).\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\left(\dfrac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x^2-4\right)}\right).\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}.\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x}{2\left(x-2\right)\left(x+2\right)}.\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{x\left(2x-1\right)}{\left(x+2\right)\left(2x-1\right)}\)
\(=\dfrac{x\left(2x-1\right)}{2x^2+4x-x-2}\)
\(=\dfrac{x\left(2x-1\right)}{2x\left(x+2\right)-\left(x+2\right)}\)
\(=\dfrac{x\left(2x-1\right)}{\left(x+2\right)\left(2x-1\right)}\)
\(=\dfrac{x}{x+2}\left(đpcm\right)\)
\(b,\)Thay \(x=1\) vào \(P\) ta có :
\(P=\dfrac{1}{1+2}=\dfrac{1}{3}\)
\(c,\dfrac{x}{x+2}< 2\)
\(\Leftrightarrow\dfrac{x-2x-4}{x+2}< 0\)
\(\Leftrightarrow-x-4< 0\)
\(\Leftrightarrow-x< 4\)
\(\Leftrightarrow x>-4\)
Vậy \(x>-4\) thì \(P< 2\)
