pt $\Leftrightarrow (x+1)(x+4)(3x+2)(3x+8)=4x^2$
$\Leftrightarrow (x+1)(3x+8)(x+4)(3x+2)=4x^2
\\\Leftrightarrow (3x^2+11x+8)(3x^2+14x+8)=4x^2$
Đặt $3x^2+11x+8=y$. Khi đó pt trở thành:
$y(y+3x)=4x^2
\\\Leftrightarrow y^2+3xy-4x^2=0
\\\Leftrightarrow y^2-xy+4xy-4x^2=0
\\\Leftrightarrow y(y-x)+4x(y-x)=0
\\\Leftrightarrow (y-x)(y+4x)=0$
Với $y-x=0\Rightarrow 3x^2+11x+8-x=0\Leftrightarrow x=\dfrac{-4}3 \ or \ x=-2$
Với $y+4x=0\Rightarrow 3x^2+11x+8+4x=0\Leftrightarrow x=\dfrac{-15\pm \sqrt{129}}6$
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