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PT $\Leftrightarrow \frac{3x}{2}\left ( \sqrt{\frac{9x^{2}}{4}+\frac{3}{4}}+1 \right )+\left ( x+\frac{1}{2} \right )\left ( \sqrt{(x+\frac{1}{2})^{2}+\frac{3}{4}} +1\right )=0
\Leftrightarrow \left ( x+\frac{1}{2} \right )\left ( \sqrt{(x+\frac{1}{2})^{2}+\frac{3}{4}}+1 \right )=-\frac{3x}{2}\left ( \sqrt{\frac{9x^{2}}{4}+\frac{3}{4}}+1 \right ) $
xét hàm f(t)=$t\left ( \sqrt{t^{2}+\frac{3}{4}}+1 \right )$ => hàm số ĐB trên R
$\Rightarrow x+\frac{1}{2}=-\frac{3x}{2} \Leftrightarrow x=-\frac{1}{5}$
vậy PT có no là $x=-\frac{1}{5}$
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