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Xét khai triển
\(\left(x+1\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1x+...+C_{2n+1}^{2n}x^{2n}+C_{2n+1}^{2n+1}x^{2n+1}\)
Cho \(x=1\) ta được:
\(2^{2n+1}=C^0_{2n+1}+C_{2n+1}^1+...+C_{2n+1}^{2n}+C_{2n+1}^{2n+1}\)
\(\Leftrightarrow2^{2n+1}=2+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n}\)
\(\Leftrightarrow2^{2n+1}-2=C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n}\)
\(\Leftrightarrow2^{10}-1=2^{2n+1}-2\Rightarrow2^{2n+1}=2^{10}+1\)
Không tồn tại n thỏa mãn yêu cầu bài toán (bạn xem lại đề bài)
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.
\(y'=4\left(x-5\right)^3\)
Câu 2:
\(\left(x^2-3y^2\right)dx+7xydy=0\)
- Với \(x=0\) là 1 nghiệm của pt đã cho
- Với \(x\ne0\)
\(\Leftrightarrow dy+\frac{1}{7}\left(\frac{x^2-3y^2}{xy}\right)dx=0\)
\(\Leftrightarrow dy+\frac{1}{7}\left(\frac{x}{y}-\frac{3y}{x}\right)dx=0\)
Đặt \(u=\frac{y}{x}\Rightarrow y=ux\Rightarrow dy=u.dx+x.du\)
\(\Leftrightarrow u.dx+x.du+\frac{1}{7}\left(\frac{1}{u}-3u\right)dx=0\)
\(\Leftrightarrow\left(\frac{4u^2+1}{7u}\right)dx=-x.du\)
\(\Leftrightarrow\frac{7u.du}{4u^2+1}+\frac{dx}{x}=0\)
\(\Leftrightarrow\frac{7}{8}.\frac{d\left(4u^2+1\right)}{4u^2+1}+\frac{dx}{x}=0\)
Lấy tích phân 2 vế:
\(\Rightarrow\frac{7}{8}\int\frac{d\left(4u^2+1\right)}{4u^2+1}+\int\frac{dx}{x}=C\)
\(\Leftrightarrow\frac{7}{8}ln\left(4u^2+1\right)+ln\left|x\right|=C\)
\(\Leftrightarrow\frac{7}{8}ln\left(\frac{4y^2}{x^2}+1\right)+ln\left|x\right|=C\)
1.
\(x_{n+2}-3x_{n+1}+2x_n=12cos\frac{n\pi}{2}+7sin\frac{n\pi}{2}\)
Xét pt thuần nhất: \(x_{n+2}-3x_{n+1}+2x_n=0\)
Pt đặc trưng: \(\lambda^2-3\lambda+2=0\Rightarrow\left[{}\begin{matrix}\lambda=1\\\lambda=2\end{matrix}\right.\)
Nghiệm của pt thuần nhất: \(\overline{x_n}=c_1+c_2.2^n\)
- Nghiệm riêng \(x_n^0\)
Do Pt đặc trưng cho nghiệm thực và các hệ số của lượng giác là hằng số bậc 0 nên nghiệm riêng có dạng: \(x_n^0=p.cos\frac{n\pi}{2}+q.sin\frac{n\pi}{2}\) với p;q là các số thực
Thay vào pt:
\(p.cos\frac{\left(n+2\right)\pi}{2}+q.sin\frac{\left(n+2\right)\pi}{2}-3pcos\frac{\left(n+1\right)\pi}{2}-3q.sin\frac{\left(n+1\right)\pi}{2}+2p.cos\frac{n\pi}{2}+2q.sin\frac{n\pi}{2}=12cos\frac{n\pi}{2}+7sin\frac{n\pi}{2}\)
\(\Leftrightarrow-p.cos\frac{n\pi}{2}-q.sin\frac{n\pi}{2}+3p.sin\frac{n\pi}{2}-3qcos\frac{n\pi}{2}+2p.cos\frac{n\pi}{2}+2q.sin\frac{n\pi}{2}=12cos\frac{n\pi}{2}+7sin\frac{n\pi}{2}\)
\(\Leftrightarrow\left(p-3q\right)cos\frac{n\pi}{2}+\left(q+3p\right)\frac{n\pi}{2}=12cos\frac{n\pi}{2}+7sin\frac{n\pi}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}p-3q=12\\3p+q=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}p=\frac{33}{10}\\q=-\frac{29}{10}\end{matrix}\right.\)
Vậy nghiệm riêng có dạng:
\(x_n^0=\frac{33}{10}.cos\frac{n\pi}{2}-\frac{29}{10}.sin\frac{n\pi}{2}\)
Nghiệm tổng quát: \(x_n=c_1+c_2.2^n+\frac{33}{10}.cos\frac{n\pi}{2}-\frac{29}{10}.sin\frac{n\pi}{2}\)
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 \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)
Để hàm liên tục tại \(x=1\)
\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow k=2\sqrt{2019}\)
2.
\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}a+b+1=0\\\lim\limits_{x\rightarrow1}\frac{2x+a}{2x}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\\frac{a+2}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=0\end{matrix}\right.\) \(\Rightarrow S=1\)
3.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)
\(\Rightarrow a+b+c=1+12+0=13\)