Đáp án: $C$
Giải thích các bước giải:
Ta có:
$\lim_{x\to2}\dfrac{\sqrt[3]{x^2+4}+\sqrt{x+2}-2x^2+x+2}{x-2}$
$=\lim_{x\to2}\dfrac{(\sqrt[3]{x^2+4}-2)+(\sqrt{x+2}-2)-(2x^2-8)+(x-2)}{x-2}$
$=\lim_{x\to2}\dfrac{\dfrac{x^2+4-2^3}{(\sqrt[3]{x^2+4})^2+2\sqrt[3]{x^2+4}+4}+\dfrac{x+2-2^2}{\sqrt{x+2}+2}-2(x^2-4)+(x-2)}{x-2}$
$=\lim_{x\to2}\dfrac{\dfrac{(x-2)(x+2)}{(\sqrt[3]{x^2+4})^2+2\sqrt[3]{x^2+4}+4}+\dfrac{x-2}{\sqrt{x+2}+2}-2(x-2)(x+2)+(x-2)}{x-2}$
$=\lim_{x\to2}\dfrac{x+2}{(\sqrt[3]{x^2+4})^2+2\sqrt[3]{x^2+4}+4}+\dfrac{1}{\sqrt{x+2}+2}-2(x+2)+1$
$=\dfrac{2+2}{(\sqrt[3]{2^2+4})^2+2\sqrt[3]{2^2+4}+4}+\dfrac{1}{\sqrt{2+2}+2}-2(2+2)+1$
$=-\dfrac{77}{12}$
Mà $\lim_{x\to2}\dfrac{\sqrt[3]{x^2+4}+\sqrt{x+2}-2x^2+x+2}{x-2}=-\dfrac mn$
$\to m=77, n=12$
$\to m-4n=29$
$\to C$
