Question

cho biểu thức

P = 1 + \(\frac{x+3}{x^2+5x+6}\) : ( \(\frac{8x^2}{4x^3-8x^2}\) - \(\frac{3x}{3x^2-12}\) - \(\frac{1}{x+2}\))

a) Rút gọn P

b) Tìm các giá trị của x để P = 0; P = 1

c) Tìm các giá trị của x để P > 0

Answer 1

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 :

\(\frac{x+4}{6}>0\Leftrightarrow x+4>0\Leftrightarrow x>-4\)