K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 6 2018

             \(D=\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\)

\(\Leftrightarrow\)\(4D=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\)

                  \(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\)

                  \(=\left(5^4-1\right)\left(5^4+1\right)\)

                  \(=5^8-1\)

\(\Leftrightarrow\)\(D=\frac{5^8-1}{4}\)

28 tháng 6 2018

có trong câu hỏi tương tự đấy bạn

1 tháng 7 2016

\(Q=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Rightarrow2Q=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=5^{32}-1\)

\(Q=\frac{5^{32}-1}{2}\)

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{5^{32}-1}{2}\)

23 tháng 9 2021

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{32}+1\right)\)

\(2P\)\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\)\(\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=5^{32}-1\)

\(P=\dfrac{5^{32}-1}{2}\)

 

20 tháng 12 2016

5^16-1 đó bạn(Cách giải:3*8=24=5^2-1 từ đó bạn áp dụng hằng đẳng thức a+1)(a-1)=a^2-1 để phân tích)

18 tháng 6 2016

Đặt \(A=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=5^{32}-1\)

Vậy \(A=\frac{5^{32}-1}{2}\)

18 tháng 6 2016

\(\frac{12.\left(5^2+1\right)\left(5^2-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{5^2-1}\)

=\(\frac{12.\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)

=\(\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

=\(\frac{\left(5^{16}-1\right)\left(5^{16+1}\right)}{2}\)

=\(\frac{5^{32}-1}{2}\)

18 tháng 5 2015

B = 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

B = 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

B = 1 - 1/7

B = 6/7 

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)