giải pt : $\sqrt{4x+6}+2=x^{2}+\sqrt[3]{x^{3}+7x^{2}+12x+6}$
$\sqrt{4x+6}+2=x^{2}+\sqrt[3]{x^{3}+7x^{2}+12x+6}$
Started By conan98md, 16-04-2014 - 19:46
#1
Posted 16-04-2014 - 19:46
#2
Posted 17-04-2014 - 23:27
giải pt : $\sqrt{4x+6}+2=x^{2}+\sqrt[3]{x^{3}+7x^{2}+12x+6}$
Giải:
$D=\left[-\frac{3}{2};+\infty \right ) $
$\Leftrightarrow\sqrt{4x+6}+2=x^{2}+\sqrt[3]{x^{3}+7x^{2}+12x+6}$
$\Leftrightarrow\sqrt{4x+6}-(x+2)=x^{2}-2 +\sqrt[3]{x^{3}+7x^{2}+12x+6}- (x+2)$
$\Leftrightarrow (x^2 -2)\left[\frac{1}{\sqrt{4x+6}+x+2}+1 + \frac{1}{ a^2+a(x+2) + (x+2)^2}\right]=0(a=\sqrt[3]{x^{3}+7x^{2}+12x+6})$
$\Leftrightarrow x= \pm\sqrt{2}\vee\frac{1}{\sqrt{4x+6}+x+2}+1 + \frac{1}{ a^2+a(x+2) + (x+2)^2}=0(!) (a=\sqrt[3]{x^{3}+7x^{2}+12x+6})$
Vậy $S= \left \{ \pm\sqrt{2} \right \}$
Edited by xxSneezixx, 17-04-2014 - 23:27.
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