1e+84937

Ta có x_n luôn dương

Ta có \(2x_n+1=\) \(2\times\dfrac{\left(2+cos\alpha\right)x_n+cos^2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}+1=\)

\(=\dfrac{6x_n+2cos^2\alpha+2-cos2\alpha}{\left(2-2cos2\alpha\right)x_n+2-cos2\alpha}\)

\(=\dfrac{6x_n+2cos^2\alpha+2sin^2a+1}{\left(2x_n+1\right)\left(1-cos2\alpha\right)+1}\)

\(=\dfrac{3\left(2x_n+1\right)}{2\sin^2\alpha\left(2x_n+1\right)+1}\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}=\dfrac{2\sin^2\alpha\left(2x_n+1\right)+1}{3\left(2x_n+1\right)}\)

\(=\dfrac{1}{3}\left(2\sin^2\alpha+\dfrac{1}{2x_n+1}\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\dfrac{1}{3}\left(\dfrac{1}{2x_n+1}-\sin^2\alpha\right)\)

\(\Rightarrow\dfrac{1}{2x_{n+1}+1}-\sin^2\alpha=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{2x_1+1}-\sin^2\alpha\right)\)

\(=\left(\dfrac{1}{3}\right)^n\left(\dfrac{1}{3}-\sin^2\alpha\right)\)

\(\Rightarrow y_n=\sum\limits^{n-1}_{i=0}\left(\dfrac{1}{3}\right)^i\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

\(=\dfrac{1-\left(\dfrac{1}{3}\right)^n}{1-\dfrac{1}{3}}\left(\dfrac{1}{3}-\sin^2\alpha\right)+n\sin^2\alpha\)

Ta có \(\overrightarrow{BH}=\overrightarrow{BC}+\overrightarrow{CH}=-\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{CA}\)

\(\overrightarrow{MH}=\dfrac{1}{2}\overrightarrow{BA}=\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CA}\right)=\dfrac{1}{2}\overrightarrow{CA}-\dfrac{1}{2}\overrightarrow{CB}\)

\(x_1=a>2;x_{n+1}=x_n^2-2,\forall n=1,2,...\)

mà \(n\rightarrow+\infty\)

\(\Rightarrow a\rightarrow+\infty\Rightarrow x_n\rightarrow+\infty\)

\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\dfrac{1}{x_n}=0\) \(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_nx_{n+1}}\right)=0\)

\(\)\(\Rightarrow\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{x_1}+\dfrac{1}{x_1x_2}+\dfrac{1}{x_1x_2x_3}+...+\dfrac{1}{x_1x_2...x_n}\right)=0\)

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Hiện tại sách là chương trình mới những vẫn không có PowerPoint bạn nha. Nên hiện tại chưa biết được khi nào sẽ có PowerPoint, nhưng ở các lớp dưới thì cũng đã học rồi á bạn.

Theo chương trình sgk mới thì lớp 7, lớp 8 có học PowerPoint. Lớp 11 thì chưa có em nhé!

\(a,-125^o=\dfrac{\pi.\left(-125\right)}{180}rad=-\dfrac{25\pi}{36}rad\\ b,42^o=\dfrac{\pi.42}{180}rad=\dfrac{7\pi}{30}rad\)

a, 25/35 \(\pi\)

b, 7/30  \(\pi\)

\(tanx=\dfrac{1}{cotx}=\dfrac{1}{\sqrt[]{2}}=\dfrac{\sqrt[]{2}}{2}\left(tanx.cotx=1\right)\)

\(1+tan^2x=\dfrac{1}{cos^2x}\Rightarrow cos^2x=\dfrac{1}{1+tan^2x}=\dfrac{1}{1+\dfrac{1}{2}}\)

\(\Rightarrow cos^2x=\dfrac{2}{3}\Rightarrow cosx=\sqrt[]{\dfrac{2}{3}}\)

\(tanx=\dfrac{sinx}{cosx}\Rightarrow sinx=tanx.cosx=\dfrac{1}{\sqrt[]{2}}.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{3}\)

\(P=\dfrac{3sinx-2cosx}{12sin^3x+4cos^3x}=\dfrac{3.\dfrac{\sqrt[]{3}}{3}-2.\dfrac{\sqrt[]{2}}{\sqrt[]{3}}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)

\(=\dfrac{\sqrt[]{3}-\dfrac{2\sqrt[]{6}}{3}}{12.\left(\dfrac{\sqrt[]{3}}{3}\right)^3+4.\left(\sqrt[]{\dfrac{2}{3}}\right)^3}\)