Question

tìm x để A = \(\dfrac{6x^2+4x}{\left(x-2\right)\left(x-3\right)}>1\)

Answer 1

Để A>1 thì A-1>0

=>\(\dfrac{6x^2+4x-x^2+5x-6}{x^2-5x+6}>0\)

=>\(\dfrac{5x^2+9x-6}{\left(x-2\right)\left(x-3\right)}>0\)

TH1: 5x^2+9x-6>0 và (x-2)(x-3)>0

=>\(\left\{{}\begin{matrix}x\in R\backslash\left[2;3\right]\\\left[{}\begin{matrix}x< \dfrac{-9-\sqrt{201}}{10}\\x>\dfrac{-9+\sqrt{201}}{10}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\in\left(-\infty;\dfrac{-9-\sqrt{201}}{10}\right)\cup\left(3;+\infty\right)\)

TH2: 5x^2+9x-6<0 và (x-2)(x-3)<0

=>\(\left\{{}\begin{matrix}2< =x< =3\\\dfrac{-9-\sqrt{201}}{10}< x< \dfrac{-9+\sqrt{201}}{10}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)