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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 \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
heo me tim gtnn gtln cua bieu thuc:asinx + bcosx (a,b la hang so,a^2+b^2=/o)? | Yahoo Hỏi & Đáp
\(\lim\limits_{x\rightarrow0}\frac{\sqrt{3x+1}-1}{x}=\lim\limits_{x\rightarrow0}\frac{3x}{x\left(\sqrt{3x+1}+1\right)}=\lim\limits_{x\rightarrow0}\frac{3}{\sqrt{3x+1}+1}=\frac{3}{2}\)
\(\Rightarrow a^2+b^2=3^2+2^2=13\)
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.
\(y'=4\left(x-5\right)^3\)
\(\lim\limits_{x\rightarrow-\infty}\frac{ax^2-4x+5}{2x^2+x+1}=\lim\limits_{x\rightarrow-\infty}\frac{a-\frac{4}{x}+\frac{5}{x^2}}{2+\frac{1}{x}+\frac{1}{x^2}}=\frac{a}{2}=4\)
\(\Rightarrow a=8\)
\(\Rightarrow\) Có đúng 1 số thực a thỏa mãn
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
Vì b \(\ne\) 0 nên tồn tại f(x) sao cho:
\(x^2+ax+2=\left(x-1\right).f\left(x\right)\)
=> 1 + a + 2 = 0
=> a = -3; f(x) = x - 2
Khi đó:
\(\lim\limits_{x\rightarrow1}\frac{x-1}{\left(\sqrt{x}+1\right)\left(x-1\right)f\left(x\right)}=b\)
\(\Leftrightarrow\lim\limits_{x\rightarrow1}\frac{1}{\left(\sqrt{x}+1\right)f\left(x\right)}=b\)
\(\Leftrightarrow b=\lim\limits_{x\rightarrow1}\frac{1}{\left(\sqrt{x}+1\right)\left(x-2\right)}=\frac{1}{2.\left(-1\right)}=-\frac{1}{2}\)
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:
\(a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
\(c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n\)
\(\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)\)
\(=...\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1\)
\(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}\)
\(=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]\)
\(=1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
