Câu 11:
\(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{3x+4}-4}{x-4}=\left(\dfrac{a}{b}\right)\)
=>\(\dfrac{a}{b}=\lim\limits_{x\rightarrow4}\left(\dfrac{3x+4-16}{\sqrt{3x+4}+4}\cdot\dfrac{1}{x-4}\right)\)
\(\Leftrightarrow\dfrac{a}{b}=\lim\limits_{x\rightarrow4}\left(\dfrac{3}{\sqrt{3x+4}+4}\right)=\dfrac{3}{\sqrt{3\cdot4+4}+4}\)
=>\(\dfrac{a}{b}=\dfrac{3}{\sqrt{16}+4}=\dfrac{3}{4+4}=\dfrac{3}{8}\)
=>a=3;b=8
3a+b=9+8=17
Đúng 2
Bình luận (0)
