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a. Áp dụng công thức L'Hospital:
\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt[3]{x+1}-\sqrt{1-x}}=\lim\limits_{x\to 0}\frac{\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}{\frac{1}{3}(x+1)^{\frac{-2}{3}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}\)
b.
\(\lim\limits_{x\to 0}(\frac{1}{x}-\frac{1}{x^2})=\lim\limits_{x\to 0}\frac{x-1}{x^2}=-\infty\)
c. Áp dụng quy tắc L'Hospital:
\(\lim\limits_{x\to +\infty}\frac{x^4-x^3+11}{2x-7}=\lim\limits_{x\to +\infty}\frac{4x^3-3x^2}{2}=+\infty \)
d.
\(\lim\limits_{x\to 5}\frac{7}{(x-1)^2}.\frac{2x+1}{2x-3}=\frac{7}{(5-1)^2}.\frac{2.5+11}{2.5-3}=\frac{11}{16}\)
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
a/ \(y'=\dfrac{\left(x^3+2\sqrt{x-1}\right)'\left(x-1\right)-\left(x-1\right)'\left(x^3+2\sqrt{x-1}\right)}{\left(x-1\right)^2}\)
\(y'=\dfrac{\left(2x^2+\dfrac{1}{\sqrt{x-1}}\right)\left(x-1\right)-x^3-2\sqrt{x-1}}{\left(x-1\right)^2}=\dfrac{x^3-2x^2-\sqrt{x-1}}{\left(x-1\right)^2}\)
b/ \(y'=\dfrac{\left(4x^3+2x-3\right)'\left(\sqrt{x^2+2}\right)-\left(\sqrt{x^2+2}\right)'\left(4x^3+2x-3\right)}{x^2+2}\)
\(y'=\dfrac{\left(12x^2+2\right)\sqrt{x^2+2}-\dfrac{x}{\sqrt{x^2+2}}\left(4x^3+2x-3\right)}{x^2+2}\) (ban tu rut gon nhe)
c/ \(y'=\dfrac{\left(x^3+x+1\right)'\left(x^3+x+1\right)}{\left|x^3+x+1\right|}=\dfrac{\left(3x^2+1\right)\left(x^3+x+1\right)}{\left|x^3+x+1\right|}\)
d/ \(y'=\dfrac{3x^2-24x^3}{2\sqrt{x^3-6x^4+7}}\)
e/ \(y'=\dfrac{\left(x^5+1\right)'\left(2-\sqrt{x^2+3}\right)-\left(x^5+1\right)\left(2-\sqrt{x^2+3}\right)'}{\left(2-\sqrt{x^2+3}\right)^2}\)
\(y'=\dfrac{5x^4\left(2-\sqrt{x^2+3}\right)+\left(x^5+1\right)\dfrac{x}{\sqrt{x^2+3}}}{\left(2-\sqrt{x^2+3}\right)^2}\)
\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)
d, Điều kiện: x > 1
\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)
e, Điều kiện: \(x>\dfrac{1}{2}\)
\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)
f, Điều kiện: x > 4
\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
Chọn B