a:

Sửa đề: \(B=\frac{5^{99}+1}{5^{100}+1}\)

Ta có: \(5A=\frac{5^{50}+5}{5^{50}+1}=\frac{5^{50}+1+4}{5^{50}+1}=1+\frac{4}{5^{50}+1}\)

\(5B=\frac{5^{100}+5}{5^{100}+1}=\frac{5^{100}+1+4}{5^{100}+1}=1+\frac{4}{5^{100}+1}\)

Ta có: \(5^{50}+1<5^{100}+1\)

=>\(\frac{4}{5^{50}+1}>\frac{4}{5^{100}+1}\)

=>\(\frac{4}{5^{50}+1}+1>\frac{4}{5^{100}+1}+1\)

=>5A>5B

=>A>B

b: \(\frac{A}{3}=\frac{3^{49}-5}{3^{49}-15}=\frac{3^{49}-15+10}{3^{49}-15}=1+\frac{10}{3^{49}-15}\)

\(\frac{B}{3}=\frac{3^{50}-5}{3^{50}-15}=\frac{3^{50}-15+10}{3^{50}-15}=1+\frac{10}{3^{50}-15}\)

Ta có: \(3^{49}-15<3^{50}-15\)

=>\(\frac{10}{3^{49}-15}>\frac{10}{3^{50}-15}\)

=>\(\frac{10}{3^{49}-15}+1>\frac{10}{3^{50}-15}+1\)

=>\(\frac{A}{3}>\frac{B}{3}\)

=>A>B

Ta có A=1+5+5²+...+5⁵⁰

=>5A=5+5²+...+5⁵¹

=>5A-A=(5+5²+...+5⁵¹)-(1+5+5²+...+5⁵⁰)

=>4A=5⁵¹-1

=>A=(5⁵¹-1) :4

Chúc bạn học giỏi!

\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)

=> \(5\text{A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\)

=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - (\(1+5+5^2+5^3+...+5^{49}+5^{50}\) )

=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - \(1-5-5^2-5^3-...-5^{49}-5^{50}\)

=> \(4\text{A}=5^{51}-1\)

=> \(A=\dfrac{5^{51}-1}{4}\)

A=1+\((5+5^2+5^3+...+5^{50})\) 5A=\(5+5^2+5^3+...+5^{51}\) ^{5A=\((1+5+5^2+...+5^{^{ }50})+5^{51}-1\) 5A=A+\(5^{51}-1\) 5A-A=\((5^{51}-1)\) -A A=\(\dfrac{5^{51-1}}{4}\)}

\(A=1+5+5^2+5^3+....+5^{49}+5^{50}\)

\(5A=5+5^2+5^3+5^4+.....+5^{50}+5^{51}\)

\(5A-A=5+5^2+5^3+5^4+......+5^{50}+5^{51}-\left(1+5+5^2+5^3+......+5^{49}+5^{50}\right)\)

\(4A=5+5^2+5^3+5^4+......+5^{50}+5^{51}-1-5-5^2-5^3-5^4-.....-5^{49}-5^{50}\)

\(4A=5^{51}-1\)

\(A=\frac{5^{51}-1}{4}\)

5A=5+5^2+5^3+........+5^51

5A-A=(5+5^2+5^3+....+5^51)-(1+5+5^2+....+5^50)

4A=5^51-1

A=5^51-1/4

bài này chỉ làm dược vậy không tính dược kết quả

 A=1+5+5^2+5^3+...+5^49+5^50

5A= 5+5^2 +...+5^51

ta co : 5A-A= 5^51 - 1 

4A= 5^51-1

=> A= 5^51-1/4