C

B

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)

a. \(y=6x^4-6x-\sqrt{7}\)

\(\Rightarrow y'=4.6.x^3-6=24x^3-6\)

b. \(y=\left(4-3x\right)\left(2x^2+3\right)\)

\(y'=-3\left(2x^2+3\right)+4x\left(4-3x\right)=-6x^2-9+16x-12x^2=-18x^2+16x-9\)

????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{8x+1}-x-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{8x-\left(x+2\right)^2}{\left(x-1\right)\left(\sqrt{8x+1}+x+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-x^2+4x-3}{\left(x-1\right)\left(\sqrt{8x+1}+x+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(3-x\right)}{\left(x-1\right)\left(\sqrt{8x+1}+x+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{3-x}{\sqrt{8x+1}+x+2}=\dfrac{2}{3+3}=\dfrac{1}{3}\)

\(f\left(1\right)=\dfrac{2m+1}{1-3.1^2}=\dfrac{2m+1}{-2}\)

Hàm liên tục tại \(x=1\) khi:

\(\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Leftrightarrow\dfrac{2m+1}{-2}=\dfrac{1}{3}\Rightarrow m=-\dfrac{5}{6}\)

no khó lắm tui lớp 5 thôi chịu

:(((((((((((

1.a

\(\lim\limits_{x\rightarrow2}\dfrac{x^3+3x^2-9x-2}{x^3-x-6}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x^2+5x+1\right)}{\left(x-2\right)\left(x^2+2x+3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+5x+1}{x^2+2x+3}=\dfrac{15}{11}\)

b.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x+3}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-x+3}{\sqrt{x^2-x+3}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1+\dfrac{3}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{3}{x^2}}-1}=\dfrac{-1}{-2}=\dfrac{1}{2}\)