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heo me tim gtnn gtln cua bieu thuc:asinx + bcosx (a,b la hang so,a^2+b^2=/o)? | Yahoo Hỏi & Đáp
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 \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)
b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)
c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)
d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)
e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)
f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)
g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)
h)
\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)
k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
a: \(-1< =cosx< =1\)
\(\Leftrightarrow-2< =2cosx< =2\)
\(\Leftrightarrow-5< =2cosx-3< =-1\)
\(f\left(x\right)_{min}=-5\) khi cos x=-1
hay \(x=\Pi+k2\Pi\)
\(f\left(x\right)_{max}=-1\) khi cos x=1
hay \(x=k2\Pi\)
b: \(-1< =sinx< =1\)
\(\Leftrightarrow-2< =2sinx< =2\)
\(\Leftrightarrow5< =2sinx+7< =9\)
\(\Leftrightarrow\sqrt{5}< =\sqrt{2sinx+7}< =3\)
\(\Leftrightarrow3\sqrt{5}< =3\sqrt{2sinx+7}< =9\)
\(f\left(x\right)_{min}=3\sqrt{5}\) khi sin x=-1
hay \(x=-\dfrac{\Pi}{2}+k2\Pi\)
\(f\left(x\right)_{max}=9\) khi sin x=1
hay \(x=\dfrac{\Pi}{2}+k2\Pi\)
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
