đặt A=1+2+2^2+2^3+2^4+2^5+2^6+2^7

2A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8

2A-A=(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)-(1+2+2^2+1^3+2^4+2^5+2^6+2^7)

A=2^8-1

A=256-1=255

255 chia hết cho 3

nên 1+2+2^2+2^3+2^4+2^5+2^6+2^7 cũng chia hết cho 3

A = 1 + 2 + 2² + 2³ + ...+ 2⁶ + 2⁷ 

= ( 1 + 2) + ( 2² +2³ ) +( 2⁴ + 2⁵ ) + ( 2⁶ + 2⁷)           ''   có tất cả 8 số chia thành 4 cặp nhé ''

=3 + 2². ( 1 + 2) +  2⁴.(1+2) + 2⁶. ( 1 + 2) 

= 3 + 2² .3 + 2⁴.3+ 2⁶ .3

= 3. ( 1 +2² + 2⁴ + 2⁶ ) chia hết cho 3.

a) Ta có : A=2+2²+2³+...+2¹⁰

                  =(2+2²)+(2³+2⁴)+...+(2⁹+2¹⁰)

                 =2(1+2)+2³(1+2)+...+2⁹(1+2)

                =2.3+2³.3+...+2⁹.3

Vì 3\(⋮\)3 nên 2.3+2³.3+...+2⁹.3\(⋮\)3

hay A\(⋮\)3

Vậy A\(⋮\)3.

b) Ta có : A=2²+2⁴+2⁶+...+2²⁰

                  =(2²+2⁴)+(2⁶+2⁷)+...+(2¹⁸+2²⁰)

                  =2²(1+2²)+2⁶(1+2²)+...+2¹⁸(1+2²)

                 =2².5+2⁶.5+...+2¹⁸.5

Vì 5\(⋮\)5 nên 2².5+2⁶.5+...+2¹⁸.5\(⋮\)5

hay A\(⋮\)5

Vậy A\(⋮\)5.

c) Ta có : A=7+7²+7³+...+7¹⁰

                  =(7+7²)+(7³+7⁴)+...+(7⁹+7¹⁰)

                  =7(1+7)+7³(1+7)+...+7⁹(1+7)

                 =7.8+7³.8+...+7⁹.8

Mà 8 chia hết cho 8 nên 7.8+7³.8+...+7⁹.8 chia hết cho 8

hay A chia hết cho 8

Vậy A chia hết cho 8.

1)Tính:

a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)

b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)

c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)

a

\(4^2.2=16.2=32\)

b\(36^2:6^2=36.36:6.6=36.36:36=36\)

c

3)

a)\(x:\left(-\frac{1}{3}\right)^5=\left(-\frac{1}{3}\right)^3\)

                      \(x=\left(-\frac{1}{3}\right)^3\cdot\left(-\frac{1}{3}\right)^5\)           

                      \(x=\left(-\frac{1}{3}\right)^8\)

b)\(\left(\frac{2}{3}\right)^{16}\cdot x=\left(\frac{2}{3}\right)^{18}\)

                      \(x=\left(\frac{2}{3}\right)^{18}:\left(\frac{2}{3}\right)^{16}\)

                      \(x=\left(\frac{2}{3}\right)^2\)

                      \(x=\frac{2^2}{3^2}\)

                      \(x=\frac{4}{9}\)

2B=1+2mũ4+2 mũ 5+.......+2 mũ 99 + 2 mũ 100

1B=2/101

Nhanh lên mọi người ơi 

2+2²+2³+....+2⁸+2⁹

=(2+2²+2³)+....+(2⁷+2⁸+2⁹)

=(2+2²+2³)+....+2⁶.(2+2²+2³)

=14+...+2⁶+14

=14.(1+.....+2⁶) \(⋮\)14

Vậy 2+2²+2³+...+2⁸+2⁹  \(⋮\)14

Chúc bn học tốt

a, 2¹.5².17 = 2.25.17 = 50.17 = 850 

b, 2² + 2³ + 2⁴ = 4 + 8 + 16 = 28

c, 2⁵.3 + 2⁴:8 + 50: 5²

= 32.3 + 16:8 + 50:25

=96 + 2 + 2

= 100

d, 11² - 10² - 3²

= 121 - 100 - 9

= 21 - 9

= 12

e, 1³ + 2³ + 3³ + 4³ + 5³

= ( 1+ 2+3+4+5)²

= 15²

= 225

A = 2 + 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰

= 2 + 2².(1 + 2 + 2²) + 2⁵.(1 + 2 + 2²) + ... + 2⁹⁸.(1 + 2 + 2²)

= 2 + 7.2² + 7.2⁵ + ... + 7.2⁹⁸)

= 2 + 7.(2² + 2⁵ + ... + 2⁹⁸)

Vậy số dư khi chia A cho 7 là 2

\(A=2+2^2+2^3+2^4+2^5+...+2^{100}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{97}+2^{98}+2^{99}\right)+2^{100}\)

\(=2\left(1+2+4\right)+2^4\left(1+2+4\right)+...+2^{97}\left(1+2+4\right)+2^{100}\)

\(=7\left(2+2^4+...+2^{97}\right)+2^{100}\)

\(Vì7⋮7=>7\left(2+2^4+..+2^{97}\right)⋮7\)

Ta có:

\(2^3\equiv1\left(mod7\right)\)

\(2^{3.33}\equiv1^{33}\left(mod7\right)\equiv1\left(mod7\right)\)

\(2^{3.33}=2^{99}=>2^{100}=2^{99}.2\equiv1.2\left(mod7\right)\equiv2\left(mod7\right)\)

\(=>2^{100}\) chia \(7\) dư \(2\) mà \(7\left(2+2^4+...+2^{97}\right)⋮7\)

\(=>A\) chia \(7\) dư \(2\)