a) điều kiện \(x\ne\pm2\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x-6}{4-x^2}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{x^2-4}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x-6}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x-6\right)}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}:\dfrac{1}{3x-2x^2-6}\)
\(A=\dfrac{1}{x-2}.\dfrac{3x-2x^2-6}{1}=\dfrac{3x-2x^2-6}{x-2}\)
b) ta có : \(3x-2x^2-6=-2x^2+3x-6=-\left(2x^2-3x+6\right)\)
\(=\left(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.\dfrac{3}{2\sqrt{2}}+\left(\dfrac{3}{2\sqrt{2}}\right)^2\right)+\dfrac{39}{8}\)
\(=\left(\sqrt{2}x-\dfrac{3}{2\sqrt{2}}\right)^2+\dfrac{39}{8}\ge\dfrac{39}{8}>0\)
\(\Rightarrow A\le0\) \(\Leftrightarrow x-2\le0\) (mà đk : \(x\ne2\) \(\Rightarrow x-2\ne0\))
vậy \(A\le0\Leftrightarrow A< 0\) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\) vậy \(x< 2\)
