Đáp án+Giải thích các bước giải:

`3/( 3x^2 - 4x + 1 ) + 13/( 3x^2 + 2x + 1 ) = 6/x`  ĐKXĐ  `x  \ne  0 ; 1 ; 1/3`

`=> 3x( 3x^2 + 2x + 1 ) + 13x( 3x^2 - 4x + 1 ) = 6( 3x^2 - 4x + 1 )( 3x^2 + 2x + 1 )`

`<=> 9x^3 + 6x^2 + 3x + 39x^3 - 52x^2 + 13x = 6( 9x^4 - 12x^3 + 3x^2 + 6x^3 - 8x^2 + 2x + 3x^2 - 4x + 1 )`

`<=> 48x^3 - 46x^2 + 16x = 6( 9x^4 - 6x^3 - 2x^2 - 2x + 1 )`

`<=> 48x^3 - 46x^2 + 16x = 54x^4 - 36x^3 - 12x^2 - 12x + 6`

`<=> 54x^4 - 84x^3 + 34x^2 - 28x + 6 = 0`

`<=> 27x^4 - 42x^3 + 17x^2 - 14x + 3 = 0`

`<=> 27x^4 + 18x^2 + 3 - 14x( 3x^2 + 1 ) - x^2 = 0`

`<=> 3( 3x^2 + 1 )^2 - 14x( 3x^2 + 1 ) - x^2 = 0`

Đặt  `3x^2 + 1 = a  ( a >= 0 )`

pt `<=> 3a^2 - 14ax - x^2 = 0`

`<=> a^2 - 2 . 7/3ax + 49/9x^2 = 58/9x^2`

`<=> ( a - 7/3 )^2 = 58/9x^2`

`<=> a = ( 7 ± \sqrt{58}x )/3`

`<=> 3x^2 + 1 = ( 7 - \sqrt{58}x )/3 ( 1 )  hoặc   3x^2 + 1 = ( 7 + \sqrt{58}x )/3  ( 2 )`

`( 1 ) <=> 9x^2 + 3 = 7 - \sqrt{58}x`

`<=> 9x^2 + \sqrt{58}x - 4 = 0`

`( a = 9 ; b = \sqrt{58} ; c = -4 )`

`Δ = b^2 - 4ac = 202`

`=> \sqrt{Δ} = \sqrt{202}`

`=> x_1 = ( - b + \sqrt{Δ} )/( 2a ) = ( -\sqrt{58} + \sqrt{202} )/18` ( KTM)

`=> x_2 = ( -b - \sqrt{Δ} )/( 2a ) = ( - \sqrt{58} - \sqrt{202} )/18`  ( KTM )

`( 2 ) <=> 9x^2 + 3 = 7 + \sqrt{58}x`

`<=> 9x^2 - \sqrt{58}x - 4 = 0`

`( a = 9 ; b = -\sqrt{58} ; c = -4 )`

`Δ = b^2 - 4ac = 202`

`=> \sqrt{Δ} = \sqrt{202}`

`=> x_1 = ( -b + \sqrt{Δ} )/( 2a ) = ( \sqrt{58} + \sqrt{202} )/18` ( KTM )

`=> x_2 = ( -b - \sqrt{Δ } )/( 2a ) = ( \sqrt{58} - \sqrt{202} )/18` ( KTM )

` Vậy  pt  có  S = ∅`