Câu 20:
\(\lim\limits_{x\rightarrow2}\dfrac{2-\sqrt{x+2}}{x^2-3x+2}=-\dfrac{a}{b}\)
=>\(-\dfrac{a}{b}=\lim\limits_{x\rightarrow2}\left(\dfrac{4-x-2}{2+\sqrt{x+2}}\cdot\dfrac{1}{\left(x-2\right)\left(x-1\right)}\right)\)
=>\(-\dfrac{a}{b}=\lim\limits_{x\rightarrow2}\left(\dfrac{-1}{\left(x-1\right)\left(2+\sqrt{x+2}\right)}\right)\)
=>\(-\dfrac{a}{b}=\dfrac{-1}{\left(2-1\right)\left(2+\sqrt{2+2}\right)}=\dfrac{-1}{2+2}=-\dfrac{1}{4}\)
=>\(\dfrac{a}{b}=\dfrac{1}{4}\)
=>a=1;b=4
2a+3b=2+12=14
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