Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)
2A = \(2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^{16}-1\right)\left(3^{16}+1\right)\)= \(3^{32}-1\)
=) A = \(\frac{3^{32}-1}{2}\)
Vì \(\frac{3^{32}-1}{2}< 3^{32}-1\)=) \(A< B\)
Câu a : - 891 / 100 = - 8 , 91
Câu b : - 21 / 4 = - 5 , 25
Câu c : 5 / 2 = 2 , 5
Câu d : 4096 / 387 = 11 ( đã làm tròn )
Câu e : - 3 / 2 = - 1 , 5
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\dfrac{3^{32}-1}{2}\)
\(\Rightarrow A.2=B=3^{32}-1\)
\(\Rightarrow k=2\)
Vậy \(k=2\) là giá trị cần tìm