`\qquad 3log_2 ^2 x -log_2 x-1=0` `(x>0)`
$⇒\left[\begin{array}{l}log_2 x=\dfrac{1+\sqrt{13}}{6}\\log_2 x=\dfrac{1-\sqrt{13}}{6}\end{array}\right.$
$⇔\left[\begin{array}{l}x=2^{\frac{1+\sqrt{13}}{6}}\\x=2^{\frac{1-\sqrt{13}}{6}}\end{array}\right.$
`=>` $a.b=2^{\frac{1+\sqrt{13}}{6}} .2^{\frac{1-\sqrt{13}}{6}}=2^{\frac{1+\sqrt{13}+1-\sqrt{13}}{6}}=2^{\frac{1}{3}}=\sqrt[3]{2}$
Đáp án $B$
