\(A=3+3^2+3^3+...+3^{100}\)

\(3A=3\left(3+3^2+3^3+...+3^{100}\right)\)

\(3A=3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

\(2A=3^{101}-3\)

\(2A+3=3^{101}=3^n\)

\(n=101\)

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)

\(\Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)

\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)

\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)

\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)

Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)

Nên:

\(x+2012=0\Rightarrow x=-2012\)

\(\left\{{}\begin{matrix}a>2\\b>2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2+m\\b=2+n\end{matrix}\right.\)

\(a+b=2+m+2+n=4+m+n\)

\(ab=\left(2+m\right)\left(2+n\right)=4+2n+2m+mn\)

Điều hiển nhiên:

\(4+2n+2m>4+m+n\)

Mà \(mn>0\)

Vậy

\(ab>a+b\Rightarrowđpcm\)

\(\)

Cách khác cách của anh Hung Nguyen

\(\sqrt{5x-x^2}+2x^2-10x+6=0\)

\(\Rightarrow\sqrt{5x-x^2-\dfrac{25}{4}+\dfrac{25}{4}}+2x^2-10x+6=0\)

\(=\sqrt{5x+\left(-x^2\right)+\dfrac{25}{4}-\dfrac{25}{4}}+2x^2-10x+6=0\)

\(\Rightarrow\sqrt{\left(-x+-\dfrac{5}{2}\right)^2-\dfrac{25}{4}}+2x^2-10x+6=0\)

\(\Rightarrow\left|-x+-\dfrac{5}{2}\right|-\dfrac{5}{2}+2x^2-10x+6=0\)

\(\Rightarrow x-\dfrac{5}{2}-\dfrac{5}{2}+2x^2-10x+6=0\)

\(\Rightarrow x-5+2x^2-10x+6=0\)

\(\Rightarrow-9x+1+2x^2=0\)

Đến đây em bí r ạ:3

\(S=\overline{abc}+\overline{bca}+\overline{cab}\)

\(S=100a+10b+c+100b+10c+a+100c+10a+b\)

\(S=\left(100a+10a+a\right)+\left(100b+10b+b\right)+\left(100c+10c+c\right)\)

\(S=111a+111b+111c\)

Vì:

\(\left\{{}\begin{matrix}111a⋮3\\111b⋮3\\111c⋮3\end{matrix}\right.\) nên: \(111a+111b+111c⋮3\) nhưng lại \(⋮̸9\)

Vậy....

\(2xy-10x-5y=-15\)

\(\Rightarrow2xy-10x-5y+15=0\)

\(\Rightarrow2xy-10x-5y+25=10\)

\(\Rightarrow2x\left(y-5\right)-5\left(y-5\right)=10\)

\(\Rightarrow\left(2x-5\right)\left(y-5\right)=10\)

\(\Rightarrow2x-5;y-5\inƯ\left(10\right)\)

\(Ư\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(2x-5\) lẻ nên:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5=1\Rightarrow2x=6\Rightarrow x=3\\y-5=10\Rightarrow y=15\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5=-1\Rightarrow2x=4\Rightarrow x=2\\y-5=-10\Rightarrow y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5=5\Rightarrow2x=10\Rightarrow x=5\\y-5=2\Rightarrow y=7\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5=-5\Rightarrow2x=0\Rightarrow x=0\\y-5=-2\Rightarrow y=3\end{matrix}\right.\end{matrix}\right.\)

Vậy...

Ta có hình vẽ:

\(\widehat{yOm}\) kề bù với \(\widehat{xOy}\) nên: \(\widehat{yOm}+\widehat{xOy}=180^o\)

\(\Rightarrow\widehat{yOm}+70^o=180^o\)

\(\Rightarrow\widehat{yOm}=180^o-70^o=110^o\)

\(\widehat{xOn}\) kề bù với \(\widehat{xOy}\) nên: \(\widehat{xOn}+\widehat{xOy}=180^o\)

\(\Rightarrow\widehat{xOn}+70^o=180^o\)

\(\Rightarrow\widehat{xOn}=180^o-70^o=110^o\)

Vì \(\widehat{yOm}=\widehat{xOn}=110^o\) và \(On\) đối \(Oy\) ;\(Om\) đối \(Ox\) nên

\(\widehat{yOm}\) và \(\widehat{xOn}\) đối đỉnh

\(Ot\) là tia phân giác của \(\widehat{yOm}\) nên:

\(\widehat{tOy}=\widehat{tOm}=\dfrac{1}{2}\widehat{yOm}=\dfrac{1}{2}110^o=55^o\)

\(Ot'\) là tia phân giác của \(\widehat{xOn}\) nên:

\(\widehat{t'Ox}=\widehat{t'On}=\dfrac{1}{2}\widehat{xOn}=\dfrac{1}{2}110^o=55^o\)

Vì \(\widehat{mOt}=\widehat{xOt'}=55^o\) và \(Om\) đối \(Ox\);\(Ot\) đối \(Ot'\)

nên: \(\widehat{mOt}\) và \(\widehat{xOt'}\) đối đỉnh