\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right).....\left(1+\frac{1}{9999}\right)\)

\(=\frac{4}{3}.\frac{9}{8}.....\frac{10000}{9999}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.....\frac{100.100}{99.101}\)

\(=\frac{2.3.....100}{1.2.....99}.\frac{2.3.....100}{3.4.....101}\)

=\(=100.\frac{2}{101}=\frac{200}{101}\)

\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right).....\left(1+\frac{1}{9999}\right)\)

    \(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}......\frac{10000}{9999}\)

     \(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{100.100}{99.101}\)

       \(=\frac{2.3.4....100}{1.2.3....99}.\frac{2.3.4.....100}{3.4.5.....101}\)

         \(=100.\frac{2}{101}\)

          \(=\frac{200}{101}\)

$n$ tiến đến đâu vậy bạn?

Câu 2:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{(n+1)-n}{n(n+1)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(\Rightarrow \lim_{n\to \infty}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})=\lim_{n\to \infty}(1-\frac{1}{n+1})=1-\lim_{n\to \infty}\frac{1}{n+1}=1-0=1\)

\(\lim\limits_{x\rightarrow-\infty}\frac{x+1}{6x-2}=\lim\limits_{x\rightarrow-\infty}\frac{1+\frac{1}{x}}{6-\frac{2}{x}}=\frac{1}{6}\)

5,

Câu 1:

$S=1+\cos ^2x+\cos ^4x+...+\cos ^{2n}x=1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n=\frac{(\cos ^2x-1)(1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n}{\cos ^2x-1}$

$=\frac{(\cos ^2x)^{n+1}-1}{\cos ^2x-1}=\frac{\cos ^{2n+2}x-1}{\sin ^2x}$

c.

ĐKXĐ: \(cosx\ne1\)

\(\Leftrightarrow cos2x-1=1-cosx\)

\(\Leftrightarrow2cos^2x-1-1=1-cosx\)

\(\Leftrightarrow2cos^2x+cosx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\left(l\right)\\cosx=-\frac{3}{2}< -1\left(l\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

d.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx\ne1\end{matrix}\right.\)

\(\Leftrightarrow cos2x=tanx-1\)

\(\Leftrightarrow cos^2x-sin^2x=\frac{sinx}{cosx}-1\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)=\frac{cosx-sinx}{-cosx}\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Leftrightarrow tanx=1\left(l\right)\\cosx+sinx=-\frac{1}{cosx}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cos^2x+sinx.cosx=-1\)

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}sin2x=-1\)

\(\Leftrightarrow cos2x+sin2x=-3\)

Do \(\left\{{}\begin{matrix}cos2x\ge-1\\sin2x\ge-1\end{matrix}\right.\) \(\Rightarrow cos2x+sin2x\ge-2>-3\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy pt đã cho vô nghiệm

a.

\(\Leftrightarrow\pi cos2x=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow cos2x=\frac{1}{2}+2k\)

Do \(-1\le cos2x\le1\Rightarrow-1\le\frac{1}{2}+2k\le1\)

\(\Rightarrow k=0\)

\(\Rightarrow cos2x=\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

b.

\(\Leftrightarrow cos4x=1\)

\(\Leftrightarrow4x=k2\pi\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

23.

\(tan^2x\ge0\Rightarrow y\le2\)

\(y_{max}=2\) khi \(tanx=0\)

\(y_{min}\) không tồn tại

24.

\(-1\le cosx\le1\Rightarrow0< 1+cosx\le2\)

\(\Rightarrow y\ge\frac{1}{2}\)

\(y_{min}=\frac{1}{2}\) khi \(cosx=1\)

\(y_{max}\) ko tồn tại

19.

\(y=\sqrt{5-\frac{1}{2}\left(2sinxcosx\right)^2}=\sqrt{5-\frac{1}{2}sin^22x}\)

\(0\le sin^22x\le1\Rightarrow\frac{3\sqrt{2}}{2}\le y\le\sqrt{5}\)

\(y_{min}=\frac{3\sqrt{2}}{2}\) khi \(sin^22x=1\)

\(y_{max}=\sqrt{5}\) khi \(sin^22x=0\)

21.

\(y=2sin^2x-\left(1-2sin^2x\right)=4sin^2x-1\)

\(0\le sin^2x\le1\Rightarrow-1\le y\le3\)

\(y_{min}=-1\) khi \(sin^2x=0\)

\(y_{max}=3\) khi \(sin^2x=1\)

Bạn tự hiểu là giới hạn khi x tiến tới 1 nhé

a/\(=lim\frac{\left(x-1\right)\left(x^{2015}+x^{2014}+...+x+1\right)}{\left(x-1\right)\left(x^{2014}+x^{2013}+...+x+1\right)}=lim\frac{x^{2015}+x^{2014}+...+x+1}{x^{2014}+x^{2013}+...+x+1}=\frac{2016}{2015}\)

b/ \(=lim\frac{\left(x-1\right)\left(x^{m-1}+x^{m-2}+...+x+1\right)}{\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x+1\right)}=lim\frac{x^{m-1}+...+1}{x^{n-1}+...+1}=\frac{m}{n}\)

Hoặc nếu bạn được sử dụng L'Hopital thì cứ việc đạo hàm tử-mẫu, lẹ hơn các trên nhiều

Làm biếng viết đủ, bạn cứ tự hiểu là giới hạn khi x tiến tới gì gì đó nhé

a/ \(lim\frac{2x.sinx.cosx}{2sin^2x}=lim\frac{cosx}{\left(\frac{sinx}{x}\right)}=1\)

b/ \(lim\frac{-x}{x\left(\sqrt{1-x}+1\right)}=lim\frac{-1}{\sqrt{1-x}+1}=-\frac{1}{2}\)

c/ \(=lim\frac{1}{x}\left(\frac{x}{x+1}\right)=lim\frac{1}{x+1}=1\)

d/ \(lim\frac{\sqrt{-x}\left(2\sqrt{-x}+1\right)}{\sqrt{-x}\left(5\sqrt{-x}-1\right)}=lim\frac{2\sqrt{-x}+1}{5\sqrt{-x}-1}=\frac{1}{-1}=-1\)

giải y như t trừ câu d t ra 2/5~ như mà ko có trong đáp án ~