Đáp án: $A$
Giải thích các bước giải:
$\lim_{x\to1}\dfrac{\sqrt{x^2+x+2}-\sqrt[3]{7x+1}}{\sqrt{2}(x-1)}$
$=\lim_{x\to1}\dfrac{(\sqrt{x^2+x+2}-2)-(\sqrt[3]{7x+1}-2)}{\sqrt{2}(x-1)}$
$=\lim_{x\to1}\dfrac{\dfrac{x^2+x+2-2^2}{\sqrt{x^2+x+2}+2}-\dfrac{7x+1-2^3}{\sqrt[3]{7x+1}^2-2\sqrt[3]{7x+1}+4)}}{\sqrt{2}(x-1)}$
$=\lim_{x\to1}\dfrac{\dfrac{(x-1)(x+2)}{\sqrt{x^2+x+2}+2}-\dfrac{7(x-1)}{\sqrt[3]{7x+1}^2+2\sqrt[3]{7x+1}+4}}{\sqrt{2}(x-1)}$
$=\lim_{x\to1}\dfrac{\dfrac{x+2}{\sqrt{x^2+x+2}+2}-\dfrac{7}{\sqrt[3]{7x+1}^2+2\sqrt[3]{7x+1}+4}}{\sqrt{2}}$
$=\dfrac{\dfrac{1+2}{\sqrt{1^2+1+2}+2}-\dfrac{7}{\sqrt[3]{7.1+1}^2+2\sqrt[3]{7.1+1}+4}}{\sqrt{2}}$
$=\dfrac{\sqrt{2}}{12}$
$\to a=1,b=12, c=0\to a+b+c=13$
