`x^2 + (4x^2)/( x + 2 )^2 = 12 ( Đk : x \ne -2 )`
`<=> x^2 - 2x (2x)/( x + 2 ) + (4x^2)/( x + 2 )^2 + 2x (2x)/( x + 2 ) = 12`
`<=> ( x - (2x)/( x + 2 ) )^2 + 2x (2x)/( x + 2 ) - 12 = 0`
`<=> ( ( x^2 + 2x - 2x )/( x + 2 ) )^2 + ( 4x^2 )/( x + 2 ) - 12 = 0`
`<=> ( x^2/( x + 2 ) )^2 + ( 4x^2 )/( x + 2 ) - 12 = 0`
Đặt `x^2/( x + 2 ) = t`
`=>` Ta có :
`( x^2/( x + 2 ) )^2 + ( 4x^2 )/( x + 2 ) - 12 = 0`
`<=> t^2 + 4t - 12 = 0`
`<=> t^2 + 6t - 2t - 12 = 0`
`<=> ( t - 2 )( t + 6 ) = 0`
`<=>` \(\left[ \begin{array}{l}t-2=0\\t+6=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}t=2\\t=-6\end{array} \right.\)
TH1 :
`t = 2`
`<=> x^2/( x + 2 ) = 2`
`<=> x^2 = 2x + 4`
`<=> x^2 - 2x - 4 = 0`
`<=> ( x^2 - 2x + 1 ) - 5 = 0`
`<=> ( x - 1 )^2 - 5 = 0`
`<=> ( x - 1 + sqrt {5} )( x - 1 - sqrt {5} ) = 0`
`<=>` \(\left[ \begin{array}{l}x - 1 + \sqrt {5}=0\\x - 1 - \sqrt {5}=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = 1 - \sqrt {5}\\x = 1 + \sqrt {5}\end{array} \right.\) `( tmđk )`
TH2 :
`t = -6`
`<=> x^2/( x + 2 ) = -6`
`<=> x^2 = -6x - 12`
`<=> x^2 + 6x + 12 = 0`
`<=> ( x^2 + 6x + 9 ) + 3 = 0`
`<=> ( x^2 + 3 )^2 + 3 = 0`
mà `( x^2 + 3 )^2 ≥ 0 ∀x`
`<=> ( x^2 + 3 )^2 + 3 ≥ 3 > 0`
`=> ( x^2 + 3 )^2 + 3 \ne 0`
`=> ( x^2 + 3 )^2 + 3 = 0 ( vô lí )`
Vậy `x ∈ { 1 + sqrt {5} ;1 - sqrt {5} )`
