\(5^x+5^{x+2}=650;5^x.26=650;5^x=25;x=2\)

\(2^x+2^{x+3}=144;2^x.9=144;2^x=16;x=4\)

\(3^{x-1}+5.3^{x-1}=162;3^{x-1}.6=162;3^{x-1}=27;x=4\)

\(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\rightarrow x-5=0\&x-5=1\) hoặc x - 5 = - 1

\(x-5=1;x=6;x-5=0;x=5;x-5=-1;x=4\)

\(\left(2^2:4\right).2^n=4;2^n=2^2;n=2\)

nhìn hoa mắt luôn mà làm là đi bệnh viện

Bài 1a:

A = 2 + 2^2 + 2^3+ ...+ 2^100

2A = 2^2 + 2^3 + ...+ 2^101

2A - A = 2^2 + 2^3 + ...+ 2^101 - 2 - 2^2 - 2^3 - ... - 2^100

A = (2^2 - 2^2) + (2^3 - 2^3) + ... + (2^100 - 2^100) + (2^101 - 2)

A = 0+ 0+ 0 + ...+ 0 + 2^101 - 2

A = 2^101 - 2

Bài 2a:

A = 7^6 + 7^5 - 7^4

A = 7^4.(7^2 + 7 - 1)

A =7^4.(49 + 7 - 1)

A =7^4.(56 - 1)

A =7^4.55

A = 7^3.(7.11).5

A = 7^3.77.5 ⋮ 77 (đpcm)

Câu 2:

Ta có: \(21^{15}=\left(3.7\right)^{15}=3^{15}.7^{15}\)

mà \(27^5.49^8=\left(3^3\right)^5.\left(7^2\right)^8=3^{3.5}.7^{2.8}=3^{15}.7^{16}\)

Vì \(15< 16\)\(\Rightarrow7^{15}< 7^{16}\)

\(\Rightarrow3^{15}.7^{15}< 3^{15}.7^{16}\)\(\Rightarrow21^{15}< 27^5.49^8\)

okee bn mh suy nghĩ đã

\(M=\left(5+5^2+5^3+.....+5^{99}\right)\)chia hết cho 31

\(M=\left(5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+....+5^{97}\left(1+5+5^2\right)\right)\)

\(M=\left(5.31+5^4.31+5^7.31+....+5^{97}.31\right)\)

\(M=31\left(5+5^4+5^7+.....+5^{97}\right)\)

=> M chia hết cho 31

B1. 2^{x + 3} + 2² = 72

=> 2^{x + 3} + 4 = 72

=> 2^{x + 3} = 72 - 4

=> 2^{x + 3} = 68

=> ko có gtri x

B2 : Ta có : A = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + ... + 2²⁰⁰¹ + 2²⁰⁰²

                      = (1 + 2) + (2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷) + ... + (2²⁰⁰⁰ + 2²⁰⁰¹ + 2²⁰⁰²)

                     = 3 + 2².(1 + 2 + 2²) + 2⁵.(1 + 2 + 2² ) + ...  + 2²⁰⁰⁰ . (1 + 2 + 2²)

                    = 3 + 2².7 + 2⁵.7 + ... + 2²⁰⁰⁰ . 7

                   = 3 + (2² + 2⁵ + .... + 2²⁰⁰⁰) . 7

=> Số dư của 7 là 3