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S= u1.u1 + u2.u2+...+un.un
S = u1.(u2 - d) + u2.(u3 - d)+...+un(un+1 - d)
S = u1.u2 + u2.u3 +...+un.un+1-d(u1+u2+...+un)
Đặt A = u2.u3 + u3.u4+...+un.un+1
3d.A = u2.u3.(u4-u1) + u3.u4.(u5-u2)+...+un.un+1.(un+2-un-1)
3d.A = u2.u3.u4 - u1.u2.u3 + u3.u4.u5 - u2.u3.u4+...+un.un+1.un+2 - un-1.un.un+1
3d.A = un.un+1.un+2 - u1.u2.u3
3d.A = (u1 + d.n - d)(u1 + d.n)(u1 + d.n + d) - u1.(u1+d).(u1+2.d)
A = [(u1 + d.n - d)(u1 + d.n)(u1 + d.n + d) - u1.(u1+d).(u1+2.d)]/(3.d)
S = A + u1.(u1 + d) + d[2.u1+(n-1).d].n/2
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\frac{\pi}{2}+k2\pi\)
Mà \(-2017\le-\frac{\pi}{2}+k2\pi\le2017\)
\(\Rightarrow\frac{-2017+\frac{\pi}{2}}{2\pi}\le k\le\frac{2017+\frac{\pi}{2}}{2\pi}\)
Do k nguyên nên \(-320\le k\le321\)
Có \(321-\left(-320\right)+1=642\) nghiệm
\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)
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.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)
\(\Rightarrow a+b+c=1+12+0=13\)
Câu 1:
$S=1+\cos ^2x+\cos ^4x+...+\cos ^{2n}x=1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n=\frac{(\cos ^2x-1)(1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n}{\cos ^2x-1}$
$=\frac{(\cos ^2x)^{n+1}-1}{\cos ^2x-1}=\frac{\cos ^{2n+2}x-1}{\sin ^2x}$
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.
\(y'=4\left(x-5\right)^3\)
Đáp án là C
Ta có S p S q = p 2 q 2
⇔ p 2 u 1 + ( p - 1 ) d q 2 u 1 + ( q - 1 ) d = p 2 q 2
⇔ q 2 u 1 + ( p - 1 ) . d = p 2 u 1 + ( q - 1 ) . d
⇔ 2 u 1 ( q - p ) + ( p - q ) d = 0
⇔ d = 2 u 1
Nếu u1 = 0 thì d = 0. Khi đó Sn = 0 với mọi n, (mâu thuẫn giả thiết).
Suy ra u 1 ≠ 0
Do đó: u 2018 u 2019 = u 1 + 2017 . 2 u 1 u 1 + 2018 . 2 u 1 = 4035 4037