Question

\(\text{Cho A=}\dfrac{x-2014}{\left(\dfrac{x-2}{x+1}-\dfrac{x+1}{x-2}\right):\left(\dfrac{x-2}{x+1}+\dfrac{x+1}{x-2}\right)}\)

\(\text{Tìm x để A}\ge0\)

Answer 1

\(A=\dfrac{x-2014}{\dfrac{x^2-4x+4-x^2-2x-1}{\left(x+1\right)\left(x-2\right)}:\dfrac{x^2-4x+4+x^2+2x+1}{\left(x+1\right)\left(x-2\right)}}\)

\(=\dfrac{x-2014}{\dfrac{-6x+3}{\left(x+1\right)\left(x-2\right)}\cdot\dfrac{\left(x+1\right)\left(x-2\right)}{2x^2-2x+5}}\)

\(=\left(x-2014\right)\cdot\dfrac{2x^2-2x+5}{-6x+3}\)

Để A>=0 thì \(\left(x-2014\right)\left(-6x+3\right)>=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-2014\right)< =0\)

=>1/2<x<=2014