So sánh hai số:
\(A=3^{32}-1\)
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
K
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21 tháng 7 2017
Áp dụng liên tục a2 - b2 = (a - b)(a + b) để biến đổi . Ta có:
A = 332 - 1 = (316 - 1)(316 + 1) = (38- 1)(38 + 1)(316 + 1) = (34 - 1)(34 + 1)(38 + 1)(316 + 2) = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) =
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) = 2.B
21 tháng 7 2017
Ta có 2B = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
2B = (34-1)(34+1)(38+1)(316+1)
2B = (38-1)(38+1)(316+1)
Tương tự ta đc:
2B = 332-1
B= 332-1/2 hay B= A/2
Vậy A>B
29 tháng 6 2017
Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy
8 tháng 7 2018
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)
KT
7 tháng 7 2018
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)
Vậy \(B< A\)
18 tháng 3 2017
Nếu đề thế này thì mình có thể làm được:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\dfrac{3^{32}-1}{2}\)
=> B>A
7 tháng 6 2017
Ta có: B=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow\) 2B= \(2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^{16}-1\right)\left(3^{16}+1\right)\)
= \(3^{32}-1\)
\(\Rightarrow\) B= \(\dfrac{3^{32}-1}{2}\)
Mà ta có A= \(3^{32}-1\)
\(\Rightarrow\) A=2B
TH
18 tháng 7 2017
Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
7 tháng 7 2023
1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)
=(3^8-1)(3^8+1)(3^16+1)
=(3^16-1)(3^16+1)
=3^32-1
2: B=(1-3^2)(1+3^2)*...*(1+3^16)
=(1-3^4)(1+3^4)(1+3^8)(1+3^16)
=1-3^32
GH
7 tháng 7 2023
1
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)
Ta có: \(A=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^8-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Vậy A = 2B