Câu hỏi của Lê Nguyễn đức Minh

Question

Nguyễn Việt Lâm · Accepted Answer

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

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

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

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

Hàm liên tục tại $x=1$ khi:

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

Câu trả lời:

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

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

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

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

Hàm liên tục tại $x=1$ khi:

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

Độ	\({18^ \circ }\)	\(\frac{{2\pi }}{9}.\frac{{180}}{\pi } = {40^ \circ }\)	\({72^ \circ }\)	\(\frac{{5\pi }}{6}.\frac{{180}}{\pi } = {150^ \circ }\)
Radian	\(18.\frac{\pi }{{180}} = \frac{\pi }{{10}}\)	\(\frac{{2\pi }}{9}\)	\(72.\frac{\pi }{{180}} = \frac{{2\pi }}{5}\)	\(\frac{{5\pi }}{6}\)

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