Sửa đề: \(P=\left(1+\dfrac{x+3}{x^2+5x+6}\right):\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
a) Ta có: \(P=\left(1+\dfrac{x+3}{x^2+5x+6}\right):\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=\left(1+\dfrac{x+3}{\left(x+3\right)\left(x+2\right)}\right):\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=\left(1+\dfrac{1}{x+2}\right):\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\dfrac{x+3}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x+4-3x-x+2}\)
\(=\dfrac{\left(x+3\right)\left(x-2\right)}{-2x+6}\)
ĐKXĐ: $x \neq -2;-3;0;2$
Sửa đề là $3x^2-12$ bạn nhé, nếu để $3x^2+12$ thì kết quả ra rất xấu, đề là $3x^2-12$ sẽ hợp logic hơn
\(=\left(1+\dfrac{x+3}{x^2+2x+3x+6}\right):\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x^2-4\right)}-\dfrac{1}{x+2}\right)\)
\(=\left(1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}\right):\left(\dfrac{2}{x-2}-\dfrac{x}{x^2-4}-\dfrac{1}{x+2}\right)\)
\(=\dfrac{x+3}{x+2}:\left(\dfrac{2.\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\dfrac{x+3}{x+2}:\dfrac{6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{6\left(x-2\right)}{x+3}\)
Vậy $P=\dfrac{6.(x-2)}{x+3}$ với $x \neq -2;-3;0;2$
b, Với \(P=0\Leftrightarrow\dfrac{6\left(x-2\right)}{x+3}=0\Leftrightarrow6.\left(x-2\right)=0\Leftrightarrow x=2\left(Loại\right)\)
\(P=1\Leftrightarrow\dfrac{6.\left(x-2\right)}{x+3}=1\Leftrightarrow6\left(x-2\right)=x+3\Leftrightarrow5x=15\Leftrightarrow x=3\left(t.m\right)\)
c, \(P>0\Leftrightarrow\dfrac{6.\left(x-2\right)}{x+3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6.\left(x-2\right)>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}6.\left(x-2\right)< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)
Kết hợp $ĐKXĐ$ có: \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)
