Đáp án: $68$
Giải thích các bước giải:
Ta có:
$\lim_{x\to2}\dfrac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}$
$=\lim_{x\to2}\dfrac{(\sqrt[3]{8x+11}-3)+(3-\sqrt{x+7})}{(x-1)(x-2)}$
$=\lim_{x\to2}\dfrac{\dfrac{8x+11-3^3}{(\sqrt[3]{8x+11})^2+3\sqrt[3]{8x+11}+9}+\dfrac{9-(x+7)}{3+\sqrt{x+7}}}{(x-1)(x-2)}$
$=\lim_{x\to2}\dfrac{\dfrac{8x-16}{(\sqrt[3]{8x+11})^2+3\sqrt[3]{8x+11}+9}+\dfrac{2-x}{3+\sqrt{x+7}}}{(x-1)(x-2)}$
$=\lim_{x\to2}\dfrac{\dfrac{8(x-2)}{(\sqrt[3]{8x+11})^2+3\sqrt[3]{8x+11}+9}-\dfrac{x-2}{3+\sqrt{x+7}}}{(x-1)(x-2)}$
$=\lim_{x\to2}\dfrac{\dfrac{8}{(\sqrt[3]{8x+11})^2+3\sqrt[3]{8x+11}+9}-\dfrac{1}{3+\sqrt{x+7}}}{x-1}$
$=\dfrac{\dfrac{8}{(\sqrt[3]{8\cdot 2+11})^2+3\sqrt[3]{8\cdot 2+11}+9}-\dfrac{1}{3+\sqrt{2+7}}}{2-1}$
$=\dfrac7{54}$
$\to m=7, n=54\to 2m+n=68$
