Đặt A = 7²⁰¹⁰ + 7²⁰⁰⁹ + ... + 7² + 7 + 1

=> 7A = 7²⁰¹¹ + 7²⁰¹⁰ + ... + 7³ + 7² + 7

Lấy 7A trừ A theo vế ta có : 

7A - A = (7²⁰¹¹ + 7²⁰¹⁰ + ... + 7³ + 7² + 7) - (7²⁰¹⁰ + 7²⁰⁰⁹ + ... + 7² + 7 + 1)

=> 6A = 7²⁰¹¹ - 1

=> A = (7²⁰¹¹ - 1) : 6

Khi đó M = 210.(7²⁰¹¹ - 1) : 6 + 35

= 35.(7²⁰¹¹ - 1) + 35

= 35.(7²⁰¹¹ - 1 + 1)

= 35.7²⁰¹¹

= 35.7.7.7²⁰⁰⁹

= 1715.7²⁰⁰⁹ \(⋮\)1715

=> M \(⋮\)1715(ĐPCM)

Ta có : 2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰

       = 2²⁰⁰⁸.(1 + 2 + 2²)

       = 2²⁰⁰⁸.(1 + 2 + 4)

       = 2²⁰⁰⁸.7 \(⋮\)7

\(\Rightarrow\)2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰ \(⋮\)10 (đpcm)

\(2^{2008}+2^{2009}+2^{2010}\)

\(=2^{2008}\left(1+2+2^2\right)\)

\(=2^{2008}.7⋮7\)

\(\Rightarrowđpcm\)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)

\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1}{2009}\)

1,

\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)

\(|x- \frac{2}{7}|=\frac{1}{7}\)

<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)

Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)

Học tốt

\(M=-\left|x-7\right|-\left(2y+4\right)^{2008}+2009\le2009\)

Dấu '=' xảy ra khi x=7 và y=-2

Vì |x-7| \(\ge\)0

(2y+4)²⁰⁰⁸\(\ge\)0

nên M\(\le\)2009

Dấu ''='' xảy ra khi \(\hept{\begin{cases}x-7=0\\2y+4=0\end{cases}< =>\hept{\begin{cases}x=7\\y=-2\end{cases}}}\)

Vậy M_max=2009 khi x=7,y=-2

\(\hept{\begin{cases}\left|x-7\right|\ge0\\\left(2y+4\right)^{2008}\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}-\left|x-7\right|\le0\\-\left(2y+4\right)^{2008}\le0\end{cases}}\)\(\Rightarrow-\left|x-7\right|-\left(2y+4\right)^{2008}\le0\)

\(\Rightarrow2009-\left|x-7\right|-\left(2y+4\right)^{2008}\le2009\)

Nên GTLN của M là 2009 đạt được khi \(\hept{\begin{cases}\left|x-7\right|=0\\\left(2y+4\right)^{2008}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=7\\y=-2\end{cases}}\)

Ta có: \(\left\{{}\begin{matrix}\left(3x-33\right)^{2008}\ge0\\\left|y-7\right|^{2009}\ge0\end{matrix}\right.\Rightarrow\left(3x-33\right)^{2008}+\left|y-7\right|^{2009}\ge0\)

Mà \(\left(3x-33\right)^{2008}+\left|y-7\right|^{2009}\le0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(3x-33\right)^{2008}=0\\\left|y-7\right|^{2009}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-33=0\\y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=7\end{matrix}\right.\)

Vậy \(x=11;y=7\)

M=\(3^{2012}-3^{2011}+3^{2010}-3^{2009}+3^{2008}\)

= \(3^{2008}.\left(3^4-3^3+3^2-3\right)\)

= \(3^{2008}.60\)

Vì \(60⋮10\) => \(3^{2008}.60⋮10\)

Hay \(3^{2012}-3^{2011}+3^{2010}-3^{2009}+3^{2008}⋮10\)

Vậy \(M⋮10\)

Chúc bạn hk tốt !!