cho d(x) = 2x^3+x-căn 2, g(x)=3x^2+x+ căn 2.

* Hướng dẫn giải

Đáp án B

Ta có:

$f\text{'}\left(x\right)={\left(2x^3+x-\sqrt{2}\right)}^{/}=6x^2+1$

$g\text{'}\left(x\right)={\left(3x^2+x+\sqrt{2}\right)}^{/}=6x+1$

$\begin{array}{l}f\text{'}\left(x\right)>g\text{'}\left(x\right)\iff 6x^2+1>6x+1\\ \iff 6x^2-6x>0\iff x\in \left(-\infty ;0\right)\cup \left(1;+\infty \right)\end{array}$