Question

\(\dfrac{4x^2+3x-1}{x^2+5x+7}>0\)

\(\dfrac{x^2-3x+2}{x^2-4x+3}>0\)

Answer 1

`[4x^2+3x-1]/[x^2+5x+7] > 0`

   Mà `x^2+5x+7 > 0 AA x`

  `=>4x^2+3x-1 > 0`

`<=>(4x-1)(x+1) > 0`

`<=>[(x < -1),(x > 1/4):}`

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`[x^2-3x+2]/[x^2-4x+3] > 0`

Xét `x^2-3x+2=0<=>[(x=2),(x=1):}`

      `x^2-4x+3=0<=>[(x=3),(x=1):}`

Đặt `[x^2-3x+2]/[x^2-4x+3]=f(x)`

BXD:

  `=>f(x) > 0<=>x in (-oo;1)uu(1;2)uu(3;+oo)`