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NV
15 tháng 3 2022

\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)

\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)

18 tháng 5 2021

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NV
21 tháng 4 2022

\(f^3\left(2-x\right)-2f^2\left(2+3x\right)+x^2g\left(x\right)+36x=0\) (1)

Thay \(x=0\Rightarrow f^3\left(2\right)-2f^2\left(2\right)=0\Rightarrow\left[{}\begin{matrix}f\left(2\right)=0\\f\left(2\right)=2\end{matrix}\right.\)

Đạo hàm 2 vế của (1):

\(\Rightarrow-3f^2\left(2-x\right).f'\left(2-x\right)-12f\left(2+3x\right).f'\left(2+3x\right)+2x.g\left(x\right)+x^2.g'\left(x\right)+36=0\)

Thay \(x=0\)

\(\Rightarrow-3f^2\left(2\right).f'\left(2\right)-12f\left(2\right).f'\left(2\right)+36=0\)

TH1: \(f\left(2\right)=0\Rightarrow36=0\) (ktm)

TH2: \(f\left(2\right)=2\)

\(\Rightarrow-3.2^2.f'\left(2\right)-12.2.f'\left(2\right)+36=0\Rightarrow f'\left(2\right)=1\)

\(\Rightarrow A=3.2+4.1=10\)

6 tháng 11 2023

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NV
13 tháng 5 2020

\(\Delta y=4\sqrt{2\left(x+\Delta x\right)-6}-4\sqrt{2x-6}=\frac{8\Delta x}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}\)

\(f'\left(x\right)=\lim\limits_{\Delta\rightarrow0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\frac{8\Delta x}{\Delta x\left(\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}\right)}\)

\(=\lim\limits_{\Delta x\rightarrow0}\frac{8}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}=\frac{8}{2\sqrt{2x-6}}=\frac{4}{\sqrt{2x-6}}\)

b/ \(f'\left(5\right)=\frac{4}{\sqrt{2.5-6}}=2\) ; \(f\left(5\right)=4\sqrt{2.5-6}=8\)

Pt tiếp tuyến: \(y=2\left(x-5\right)+8=2x-2\)

c/ \(f'\left(x\right)>4\Leftrightarrow\frac{4}{\sqrt{2x-6}}>4\Leftrightarrow\frac{1}{\sqrt{2x-6}}>1\)

\(\Leftrightarrow\sqrt{2x-6}< 1\Leftrightarrow2x-6< 1\Rightarrow x< \frac{7}{2}\)

\(\Rightarrow3< x< \frac{7}{2}\)

NV
19 tháng 6 2019

\(sin\left(x+\frac{\pi}{6}\right)=1\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\Rightarrow x=\frac{\pi}{3}+k2\pi\)