Đặt \(A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}\)

=>\(25A=5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(A+25A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}+5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(26A=5-\frac{1}{5^{99}}=\frac{5^{100}-1}{5^{99}}\)

=>\(A=\frac{5^{100}-1}{5^{99}\cdot26}\)

rối quá :)

B = (-5)⁰ + 5¹ + (-5)² + 5³ + ... + (-5)²⁰¹⁶ + 5²⁰¹⁷

B = 1 + 5¹ + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷

5B = 5 + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷

5B - B = (5 + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷) - (1 + 5¹ + 5² + 5³ + ... + 5²⁰¹⁶ + 5²⁰¹⁷)

   4B    =    5²⁰¹⁷                                          -       1

   B      =   \(\dfrac{5^{2017}-1}{4}\)

hbewjfewi

Câu 3 = (5 mũ 51 - 1) : 4

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

A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 

2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100

2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - (  2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 ) 

A = 2^100 - 2^3 

B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 

5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 

5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )

4B = 5^51 - 1 

B = 5^51 - 1 / 4

\(\frac{5^4\cdot7^3\cdot a^5\cdot b^6}{5^3\cdot7a^3\cdot b^5}\)

\(=\frac{5^4}{5^3}\cdot\frac{7^3}{7}\cdot\frac{a^5}{a^3}\cdot\frac{b^6}{b^5}\)

\(=5\cdot7^2\cdot a^2\cdot b=245a^2b\)

aaaaaaaaaaaaaaa

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaa

đo jgfhjgh dfj hdfh vhfcvkjgb