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$x^{2}-8x+16+3x+4-8\sqrt{3x+4}+16=0$
$\Leftrightarrow (x-4)^{2}+(\sqrt{3x+4}-4)^{2}=0$
$\Leftrightarrow \left\{\begin{matrix} x-4=0 & \ & \\ \sqrt{3x+4}=4 & & \end{matrix}\right. \Leftrightarrow x=4$
tranphamminhnhut2403
In Topic: APMO 1998
17-11-2016 - 23:10
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