Question

P=\(\left(\dfrac{x^2-3x}{x^2-9}-1\right):\left(\dfrac{9-x^2}{x^2+x+6}-\dfrac{x-3}{2-x}-\dfrac{x-2}{x+3}\right)\)

b) Rút gọn P. Tìm P với x thỏa mãn x³ -4x=0

Answer 1

\(b,P=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right]:\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\left(x\ne\pm3;x\ne2\right)\\ P=\left(\dfrac{x}{x+3}-1\right)\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2+x^2-9-\left(x-2\right)^2}\\ P=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{-\left(x-2\right)^2}\\ P=\dfrac{-3}{-\left(x-2\right)}=\dfrac{3}{x-2}\)

Với \(x^3-4x=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\\x=-2\end{matrix}\right.\)

Với \(x=0\Leftrightarrow P=\dfrac{3}{0-2}=-\dfrac{3}{2}\)

Với \(x=-2\Leftrightarrow P=\dfrac{3}{-2-2}=-\dfrac{3}{4}\)