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C

- Hướng dẫn giải

Đáp án C

Điều kiện $x>0,x\ne 1$

Đặt $t={\log }_x\frac{22}{3}$

$\Rightarrow \left(\sqrt{2t^2-2t+5}+\sqrt{2t^2-4t-+4}-\sqrt{13}\right)\left(26x^6-2x^5+27x^4-2x^3+1997x^2+2016\right)\le 0\left(1\right)$

Ta có

$\sqrt{t^2-t-\frac{5}{2}}+\sqrt{t^2-2t+2}=\sqrt{{\left(t-\frac{1}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2}+\sqrt{{\left(1-t\right)}^2+1}\ge \sqrt{{\left(t-\frac{1}{2}+1-t\right)}^2+{\left(\frac{3}{2}+1\right)}^2}=\sqrt{\frac{13}{2}}$

Suy ra $\sqrt{2t^2-2t+5}+\sqrt{2t^2-4t-+4}-\sqrt{13}\ge 0$

Có $26x^6-2x^5+27x^4-2x^3+1997x^2+2016=23x^6+\left(x^6-2x^5+x^4\right)+\left(x^4-2x^3+x^2\right)+25x^4+1996x^2+2016$

$=23x^6+25x^5+1996x^2+{\left(x^3-x^2\right)}^2+\left(x^2-x^2\right)+2016>0$

Suy ra $\left(\sqrt{2t^2-2t+5}+\sqrt{2t^2-4t-+4}-\sqrt{13}\right)\left(26x^6-2x^5+27x^4-2x^3+1997x^2+2016\right)\ge 0$

Suy ra $\left(1\right)\iff \sqrt{2t^2-2t+5}+\sqrt{2t^2-4t+4}-\sqrt{13}=0\iff \frac{t-\frac{1}{2}}{1-t}=\frac{3}{2}\Rightarrow t=\frac{4}{5}$

Suy ra ${\log }_x\frac{22}{3}=\frac{4}{5}\Rightarrow x={\left(\frac{22}{3}\right)}^{\frac{5}{4}}\approx 12,1$