Sao bạn viết mũ đc vậy

\(A=1+5+5^2+5^3+...+5^n\)

=> \(5A=5+5^2+5^3+5^4+...+5^{n+1}\)

=> \(5A-A=\left(5+5^2+5^3+...+5^{n+1}\right)-\left(1+5+5^2+...+5^n\right)\)

=> \(4A=5^{n+1}-1\)

=> \(4A+1=5^{n+1}\)

mà \(4A+1=625\)

=> \(5^{n+1}=625=5^4\)

=> \(n+1=4\)

=> \(n=3\)

\(A=756\)

A = 1 + 5 + 125 + 625

A = 756

\(A=\sqrt{625}-\dfrac{1}{\sqrt{5}}=25-\dfrac{1}{\sqrt{5}}\)

\(B=\sqrt{576}-\dfrac{1}{\sqrt{6}}+1=24-\dfrac{1}{\sqrt{6}}+1=25-\dfrac{1}{\sqrt{6}}.\)

Vì \(\sqrt{5}< \sqrt{6}\) nên \(\dfrac{1}{\sqrt{5}}>\dfrac{1}{\sqrt{6}}.\)

Từ (1), (2) và (3) suy ra \(A< B.\)

B<A

a: \(A=1+\frac15+\frac{1}{25}+\cdots+\frac{1}{78125}\)

=>\(A=1+\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^7}\)

=>\(5\times A=5+1+\frac15+\cdots+\frac{1}{5^6}\)

=>\(5\times A-A=5+1+\frac15+\cdots+\frac{1}{5^6}-1-\frac15-\cdots-\frac{1}{5^7}\)

=>\(4\times A=5-\frac{1}{5^7}=\frac{5^8-1}{5^7}\)

=>\(A=\frac{5^8-1}{4\times5^7}\)

b:Sửa đề: \(B=\frac13+\frac{1}{12}+\cdots+\frac{1}{49152}\)

=>\(B=\frac13+\frac{1}{3\times4}+\frac{1}{3\times4^2}+\cdots+\frac{1}{3\times4^7}\)

=>\(4\times B=\frac43+\frac13+\frac{1}{3\times4}+\cdots+\frac{1}{3\times4^6}\)

=>\(4\times B-B=\frac43+\frac13+\frac{1}{3\times4}+\cdots+\frac{1}{3\times4^6}-\frac13-\frac{1}{3\times4}-\frac{1}{3\times4^2}-\cdots-\frac{1}{3\times4^7}\)

=>\(3\times B=\frac43-\frac{1}{3\times4^7}=\frac{4^8-1}{3\times4^7}\)

=>\(B=\frac{4^8-1}{9\times4^7}\)

c: \(C=\frac53+\frac56+\frac{5}{12}+\frac{5}{24}+\cdots+\frac{5}{192}+\frac{5}{384}\)

=>\(2\times C=\frac{10}{3}+\frac53+\frac56+\cdots+\frac{5}{96}+\frac{5}{192}\)

=>\(2\times C-C=\frac{10}{3}+\frac53+\frac56+\cdots+\frac{5}{96}+\frac{5}{192}-\frac53-\frac56-\cdots-\frac{5}{192}-\frac{5}{384}\)

=>\(C=\frac{10}{3}-\frac{5}{384}=\frac{1280}{384}-\frac{5}{384}=\frac{1275}{384}\)