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@Nguyễn Việt Lâm giúp em với

\(\dfrac{1}{u_n-1}=\dfrac{1}{\dfrac{2^n-5^n}{2^n+5^n}-1}=\dfrac{2^n+5^n}{-2.5^n}=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^n+1\right]\)

\(\Rightarrow S_n=-\dfrac{1}{2}\left[\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n+n\right]\)

Lại có: \(\left(\dfrac{2}{5}\right)^1+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n=\dfrac{2}{5}.\dfrac{1-\left(\dfrac{2}{5}\right)^n}{1-\dfrac{2}{5}}=\dfrac{2}{3}\left[1-\left(\dfrac{2}{5}\right)^n\right]\)

\(\Rightarrow S_n=-\dfrac{1}{2}\left[\dfrac{2}{3}-\dfrac{2}{3}\left(\dfrac{2}{5}\right)^n+n\right]=...\)

\(2sinx+2\sqrt{3}cosx-\sqrt{3}sin2x+cos2x=\sqrt{3}cosx+cos2x-2sinx+2\)

\(\Leftrightarrow4sinx+\sqrt{3}cosx-2\sqrt{3}sinx.cosx-2=0\)

\(\Leftrightarrow-2sinx\left(\sqrt{3}cosx-2\right)+\sqrt{3}cosx-2=0\)

\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{3}cosx-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=\dfrac{2}{\sqrt{3}}>1\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Leftrightarrow cos10x+cos2x-2cos4x.sinx=0\)

\(\Leftrightarrow2cos6x.cos4x-2cos4x.sinx=0\)

\(\Leftrightarrow cos4x\left(cos6x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos6x=sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos6x=cos\left(\dfrac{\pi}{2}-x\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

Xét t/g AHD vuông tại H có 

\(\widehat{HAD}+\widehat{BDA}=90^o\) (t/c)

=> \(\widehat{DAC}+\widehat{BDA}=90^o\)

Mà \(\widehat{DAC}+\widehat{DAB}=\widehat{BAC}=90^o\)

=> \(\widehat{BAD}=\widehat{BDA}\)

=> t/g ABD cân tại B

\(\left(5n-8\right)⋮\left(n-3\right)\\ \Rightarrow\left[5\left(n-3\right)+7\right]⋮\left(n-3\right)\\ \left[5\left(n-3\right)\right]⋮\left(n-3\right)\\ \Rightarrow7⋮\left(n-3\right)\\ \Rightarrow\left(n-3\right)\inƯ\left(7\right)\\ \Rightarrow\left(n-3\right)\in\left\{-7;-1;1;7\right\}\\ \Rightarrow n\in\left\{-4;2;4;10\right\}\)

Vậy \(n\in\left\{-4;2;4;10\right\}\)

63 và 36