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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.
\(y'=4\left(x-5\right)^3\)
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25.
H là hình chiếu của S lên (ABC)
Do \(SA=SB=SC\Rightarrow HA=HB=HC\)
\(\Rightarrow\) H là tâm đường tròn ngoại tiếp tam giác ABC
26.
\(\left\{{}\begin{matrix}AB\perp BC\\AB\perp CD\end{matrix}\right.\) \(\Rightarrow AB\perp\left(BCD\right)\) \(\Rightarrow AB\perp BD\)
\(\Rightarrow\Delta ABD\) vuông tại B
Pitago tam giác vuông BCD (vuông tại C):
\(BC^2+CD^2=BD^2\Rightarrow BD^2=b^2+c^2\)
Pitago tam giác vuông ABD:
\(AD^2=AB^2+BC^2=a^2+b^2+c^2\)
\(\Rightarrow AD=\sqrt{a^2+b^2+c^2}\)
23.
Gọi H là chân đường cao hạ từ S xuống BC
\(\Rightarrow BH=SB.cos30^0=3a\) ; \(SH=SB.sin30^0=a\sqrt{3}\) ; \(CH=4a-3a=a\)
\(\Rightarrow BC=4HC\Rightarrow d\left(B;\left(SAC\right)\right)=4d\left(H;\left(SAC\right)\right)\)
Từ H kẻ \(HE\perp AC\) ; từ H kẻ \(HF\perp SE\Rightarrow HF\perp\left(SAC\right)\)
\(\Rightarrow HF=d\left(H;\left(SAC\right)\right)\)
\(HE=CH.sinC=\frac{CH.AB}{AC}=\frac{a.3a}{5a}=\frac{3a}{5}\)
\(\frac{1}{HF^2}=\frac{1}{HE^2}+\frac{1}{SH^2}\Rightarrow HF=\frac{HE.SH}{\sqrt{HE^2+SH^2}}=\frac{3a\sqrt{7}}{14}\)
\(\Rightarrow d\left(B;\left(SAC\right)\right)=4HF=\frac{6a\sqrt{7}}{7}\)
24.
\(SA=SC\Rightarrow SO\perp AC\)
\(SB=SD\Rightarrow SO\perp BD\)
\(\Rightarrow SO\perp\left(ABCD\right)\)
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\(\lim\limits_{x\rightarrow1}\frac{x^3-4x^2+3}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-3x-3\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-3x-3}{x+1}=\frac{1-3-3}{2}=-\frac{5}{2}\)
Để hàm số liên tục tại x=1
\(\Leftrightarrow a+\frac{5}{2}=-\frac{5}{2}\Rightarrow a=-5\)
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3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
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\(y=\sqrt{5}\left(\frac{2}{\sqrt{5}}sinx+\frac{1}{\sqrt{5}}cosx\right)=\sqrt{5}.sin\left(x+a\right)\)
Do \(-1\le sina\le1\)
\(\Rightarrow-\sqrt{5}\le y\le\sqrt{5}\)
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a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
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Bài 1:
a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)
b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)
c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)
\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)
Bài 2:
\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
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Bài 1:
\(a=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(b=\frac{1-5+1}{0}=\frac{-3}{0}=-\infty\)
\(c=\lim\limits_{x\rightarrow1}\frac{x\left(1+2x\right)\left(1+3x\right)+2x\left(1+3x\right)+3x}{x}=\lim\limits_{x\rightarrow1}\left[\left(1+2x\right)\left(1+3x\right)+2\left(1+3x\right)+3\right]=1+2+3=6\)
\(d=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-1}{5x^4+2x}=\frac{4}{0}=+\infty\)
Bài 2:
\(a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
\(c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n\)
\(\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)\)
\(=...\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1\)
\(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}\)
\(=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]\)
\(=1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
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\(K=\lim\limits n\left(\sqrt[3]{1+\frac{1}{n}-\frac{1}{n^3}}-1+3\left(2-\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}\right)\right)\)
\(=\lim\limits n\left[\frac{\frac{1}{n}-\frac{1}{n^3}}{\sqrt[3]{\left(1+\frac{1}{n}-\frac{1}{n^3}\right)^2}+\sqrt[3]{1+\frac{1}{n}-\frac{1}{n^3}}+1}-\frac{3\left(\frac{1}{n}+\frac{1}{n^2}\right)}{2+\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}}\right]\)
\(=\lim\limits\left[\frac{1-\frac{1}{n^2}}{\sqrt[3]{\left(1+\frac{1}{n}-\frac{1}{n^3}\right)^2}+\sqrt[3]{1+\frac{1}{n}-\frac{1}{n^3}}+1}-\frac{3\left(1+\frac{1}{n}\right)}{2+\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}}\right]\)
\(=\frac{1}{1+1+1}-\frac{3}{2+2}=-\frac{5}{12}\)