cái dòng thứ 2 sao ra đc hay vậy??

A=75(4²⁰⁰⁴+4²⁰⁰³+...+4²+4+1)+25

=25.[3.(4²⁰⁰⁴+4²⁰⁰³+...+4²+4+1)+1]

=25.(3.4²⁰⁰⁴+3.4²⁰⁰³+...+3.4²+3.4+3+1)

=25.(3.4²⁰⁰⁴+3.4²⁰⁰³+...+3.4²+3.4+4)

=25.4.(3.4²⁰⁰³+3.4²⁰⁰²+...+3.4+3+1)

=100.(3.4²⁰⁰³+3.4²⁰⁰²+...+3.4+3+1)chia hết cho 100

=>dpcm

Bài 1: Giải:

Đặt \(B=4^{2004}+4^{2003}+...+4^2+4+1\)

\(\Rightarrow4B=4\left(4^{2004}+4^{2003}+...+4^2+4+1\right)\)

\(=4^{2005}+4^{2004}+...+4^3+4^2+4\)

\(\Rightarrow4B-B=\left(4^{2005}+4^{2004}+...+4^2+4\right)-\left(4^{2004}+4^{2003}+...+4+1\right)\)

\(\Rightarrow3B=4^{2005}-1\Rightarrow B=\dfrac{4^{2005}-1}{3}\)

Do đó:

\(A=75.\dfrac{4^{2005}-1}{3}+25=25\left(4^{2005}-1+1\right)\)

\(=25.4^{2005}=25.4.4^{2004}=100.4^{2004}⋮100\) (Đpcm)

đề thiếu rồi nek bạn

\(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)

\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)

\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)

P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r

tui lớp 8 ko bt làm :)

trời ơi cíu tui

B=4^2004+4^2003+...+4^2+4+1

4B = 4^2005+4^2004+...+4^2+4
=> 4B-B = (4^2005+4^2004+...4^3+4^2+4) - (4^2004+4^2003+...+4^2+4+1)
=> 3B = 4^2005 - 1 => B = (4^2005 - 1)/3
=> A = 75 (4^2005 - 1)/3 +25
= 25 (4^2005 -1) +25
= 25 x 4 ^ 2005
= 25 x 4 x 4 ^ 2004 = 100 x4 ^ 2004 chia hết cho 100 ( Vì 100 chia hết cho 100 )