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Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)

a, A=31+32+33+...+32006
3A=32+33+...+32006+32007
3A-A=(32+33+...+32006+32007)-(31+32+33+...+32006)
2A=32007-3
A=(32007-3)/2
b, 2A=32007-3
2A+3=32007
Hay 3x=32007
=>x=2007

Mình làm ngắn gọn nhé.
\(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)
\(\Rightarrow A=2^{51}-1\)
\(B=1+3+...+3^{66}\)
\(3B=3+3^2+...+3^{67}\)
\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)
\(2B=3^{67}-1\)
\(B=\frac{3^{67}-1}{2}\)

a. \(12^2.3^2.2^3=2^4.3^2.3^2.2^3=2^7.3^4\)
b. \(8^3.3^2.6^3=2^9.3^2.2^3.3^3=2^{12}.3^5\)
c. \(5^{32}.5^2=5^{34}\)
d. \(100^6.2^3=\left(2^2.5^2\right)^6.2^3=2^8.5^8.2^3=2^{11}.5^8\)
e. \(100^2:10^2:5^2=\left(10.5.2\right)^2:10^2:5^2=2^2\)
f. \(121^3-11^2=11^6-11^2=11^2\left(11^4-1\right)\)
\(M=3+3^2+3^3+...+3^{100}\)
\(2M=3^2+3^3+...+3^{101}\)
\(2M-M=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(M=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\)
\(M=3^{101}-3\)
a) Ta có :
M = 3 + 32 + 33 + ... + 3100
=> 3M = 32 + 33 + 34 + ... + 3101
=> 3M - M = ( 32 + 33 + 34 + ... + 3101 ) - ( 3 + 32 + 33 + ... + 3100 )
=> 2M = 3101 - 3
=> M = \(\frac{3^{101}-3}{2}\)
b) Ta có :
M = 3 + 32 + 33 + ... + 3100
=> 3M = 32 + 33 + 34 + ... + 3101
=> 3M - M = ( 32 + 33 + 34 + ... + 3101 ) - ( 3 + 32 + 33 + ... + 3100 )
=> 2M = 3101 - 3
=> 2M + 3 = 3101 - 3 + 3 = 3101