\(10^6\) tận cùng là 0 \(=>10^6+2\) tận cùng là 2 \(=>10^6+2\) chia hết cho 2

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)

\(A=2^1+2^2+2^3+...+2^{2010}\)

\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)

A=21+22+23+...+22010

=(21+22)+(23+24)+...+(22009+22010)=(21+22)+(23+24)+...+(22009+22010)

=2(1+2)+23(1+2)+...+22009(1+2)=2(1+2)+23(1+2)+...+22009(1+2)

=3(2+23+...+22009)⋮3=3(2+23+...+22009)⋮3

�=21+22+23+...+22010A=21+22+23+...+2...

\(A=2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)

\(=3\left(2+2^3+...+2^9\right)⋮3\)

\(A=2+2^2+2^3+...+2^{10}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)

\(=\left(2+2^6\right).31⋮31\)

sao ko dung f(x) ma viet

\(a=2+2^2+2^3+2^4+2^5+2^6+2^7+2^9+2^{10}\)

a=\(\left(2+2^2\right)+2^2.\left(2+2^2\right)+..+2^8\left(2+2^2\right)\)

a=\(\left(2+2^2\right).\left(1+2^2+..+2^8\right)\)

a=\(6.\left(1+2^2+2^4+2^6+2^8\right)\)

chia het cho 3

\(3+3^2+3^3+...+3^{60}\\ =\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\\ =\left(1+3\right)\left(3+3^3+...+3^{59}\right)\\ =4\left(3+3^3+...+3^{59}\right)⋮4\\ 3+3^2+3^3+...+3^{60}\\ =\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ =3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)\\ =13\left(3+3^4+...+3^{58}\right)⋮13\)

thanks

a,18³:9³=(18-9)³=729

b,còn lại tự làm ,làm mẫu đến đây thôi

l.i.k.e cho mình nha Nguyen vu dang nguyen

18³: 9³=8

C) 213

D)0

co ai giai dc ko cau 60 % .....

$\textbf{Ta có:}$

$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$

$=\dfrac{-25-45}{125+225}$

$=\dfrac{-70}{350}$

$=-\dfrac15.$

Và

$B=\dfrac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}$

$=\dfrac{2^{12}3^{10}+2^{12}3^{10}\cdot5}{2^{11}3^{11}(2\cdot3-1)}$

$=\dfrac{2^{12}3^{10}(1+5)}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^{12}3^{10}\cdot6}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^2}{5}$

$=\dfrac45.$

=> $M=B-A$$=\dfrac45-\left(-\dfrac15\right)$

$=\dfrac55$$=1.$

2³+2⁴+2⁵+2⁶+2⁹⁸+2007⁰ đề thế hả bạn

gọi tổng đó là A:

A=1+2²+2³+2⁴+...+2⁹⁸

2A=2²+2³+2⁴+...+2⁹⁸+2⁹⁹

2A-A=2⁹⁹-1

A=2⁹⁹-1