Chủ đề này có 1 trả lời

[buingoctu]

Giải PT sau:

$\sqrt{\sqrt{3}-x}=x\sqrt{\sqrt{3}+x}$

 

[MarkGot7]

ĐK: $0\leq x\leq 3$

 $\sqrt{\sqrt{3}-x}= x\sqrt{\sqrt{3}+x}$

$\Leftrightarrow \sqrt{3}-x= x^{2}\sqrt{3}+x^{3}$

$\Leftrightarrow x^{2}\sqrt{3}+x^{3}+x-\sqrt{3}=0$

$\Leftrightarrow x^{3}+3\frac{1}{\sqrt{3}} x^{2}+3\frac{1}{3}x+\frac{1}{3\sqrt{3}}= \frac{10}{3\sqrt{3}}$

$\Leftrightarrow \left ( x+\frac{1}{\sqrt{3}} \right )^{3}=\frac{10}{3\sqrt{3}}$

$\Leftrightarrow x+ \frac{1}{\sqrt{3}}= \sqrt[3]{\frac{10}{3\sqrt{3}}}$

$\Leftrightarrow x+\frac{1}{\sqrt{3}}= \frac{\sqrt[3]{10}}{\sqrt[3]{(\sqrt{3})^{3}}}$

$\Leftrightarrow x=\frac{\sqrt[3]{10}-1}{\sqrt{3}}$ ( t/m)