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đkxđ : x ≠ 2
a) Ta có :
\(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}\div\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=1+\frac{1}{x+2}\div\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=1+\frac{1}{x+2}\div\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}=1+\frac{1}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{6}\)
\(=1+\frac{x-2}{6}=\frac{x+4}{6}\)
b) Để P = 0 thì :
\(\frac{x+4}{6}=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
Để P = 1 thì :
\(\frac{x+4}{6}=1\Leftrightarrow x+4=6\Leftrightarrow x=2\)
c) Để P > 0 thì :
\(\frac{x+4}{6}>0\Leftrightarrow x+4>0\Leftrightarrow x>-4\)
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??