Question

Answer 1

\(f\left(2\right)=2m+n\cdot2+1=2m+2n+1\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}2m+nx+1=2m+2n+1\)

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{\sqrt{x+2}-2+m\left(x-2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2^+}\dfrac{\dfrac{x-2}{\sqrt{x+2}+2}+m\left(x-2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2^+}\dfrac{1}{\sqrt{x+2}+2}+m=\dfrac{1}{4}+m\)

Để hàm số có đạo hàm tại x=2 thì \(m+\dfrac{1}{4}=2m+2n+1\)

=>\(m-2m-2n=\dfrac{3}{4}\)

=>\(-m-2n=\dfrac{3}{4}\)

=>\(m+2n=-\dfrac{3}{4}\)