a) \(P=\dfrac{x^2+2}{x^2+2x-3}:\left(\dfrac{3+x}{3x-3}-\dfrac{1+x}{2x+6}+\dfrac{x^2-27}{6x^2+12x-18}\right)\left(x\ne1;x\ne-3;x\ne\dfrac{1}{2}\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x+3\right)}:\left[\dfrac{3+x}{3\left(x-1\right)}-\dfrac{1+x}{2\left(x+3\right)}+\dfrac{x^2-27}{6\left(x^2+2x-3\right)}\right]\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x+3\right)}:\left[\dfrac{2\left(x+3\right)^2}{2\cdot3\left(x-1\right)\left(x+3\right)}-\dfrac{3\left(x+1\right)\left(x-1\right)}{3\cdot2\left(x+3\right)\left(x-1\right)}+\dfrac{x^2-27}{6\left(x-1\right)\left(x+3\right)}\right]\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x+3\right)}:\dfrac{2\left(x^2+6x+9\right)-3\left(x^2-1\right)+x^2-27}{6\left(x-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x+3\right)}:\dfrac{2x^2+12x+18-3x^2+3+x^2-27}{6\left(x-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x+3\right)}:\dfrac{12x-6}{6\left(x-1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+2}{\left(x+1\right)\left(x-3\right)}\cdot\dfrac{6\left(x-1\right)\left(x+3\right)}{6\left(2x-1\right)}\)
\(=\dfrac{x^2+2}{2x-1}\)
b) Để \(P\) là số âm thì: \(\dfrac{x^2+2}{2x-1}< 0\)
\(\Rightarrow2x-1< 0\) (vì \(x^2+2>0\))
\(\Leftrightarrow2x< 1\Leftrightarrow x< \dfrac{1}{2}\)
Kết hợp với điều kiên xác định của \(x\), ta được: \(x< \dfrac{1}{2};x\ne-3\)
\(\text{#}Toru\)
