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10 tháng 11 2017
\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)
\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)
\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)
11 tháng 7 2021
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
DG
14 tháng 7 2016
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=3(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)=2^32-1
TN
1
13 tháng 8 2021
Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Leftrightarrow A< B\)
DT
30 tháng 6 2018
1) ta có \(\left(x+y\right)^2=x^2+2xy+y^2.\)
\(=\left(x^2+y^2\right)+2xy\)
\(=20+2.8\)(theo giả thiết x^2+y^2=20 , xy=8)
\(=36\)
Vậy với x^2+y^2=20, xy=8 thì (x+y)^2=36
2) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Rightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^2\right)^2-1^2\right]\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^4\right)^2-1^2\right]\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left[\left(2^8\right)^2-1^2\right]\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}\right)^2-1^2\)
\(\Leftrightarrow3M=2^{32}-1\)
\(\Rightarrow M=\frac{2^{32}-1}{3}\)
RÚT GỌN BIỂU THỨC N BẠN LÀM TƯƠNG TỰ NHA
\(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Rightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(...\)
\(...\)
Kết quả rút gọn \(N=\frac{7^{32}-1}{3}\)
AH
Akai Haruma
Giáo viên
4 tháng 8 2021
Bài 2:
a.
$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2=2^2=4$
b.
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$
c.
$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$
$\Rightarrow C=\frac{5^{32}-1}{2}$
VT
2
19 tháng 7 2015
A = (2 - 1)(2 + 1)(2^2 + 1 )(2^4 + 1 ) (2^8 + 1)(2^16 + 1) ( nhân vói 2 - 1 = 1 Gía không thay dổi)
A = ( 2 ^2 - 1 )(2^2 + 1 )(2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = ( 2^4 - 1 )(2^4 + 1)(2^8 + 1)(2^16 + 1)
A = (2^8 - 1)(2^8 + 1)(2^16 + 1)
A = (2^16 - 1)(2^16 + 1 )
A = 2^32 - 1 <2^32 = B
VẬy A < B
A=65536 > B=65535 do:
A=\(2^{16}\)
=65536
B=\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=3x5x17x257
=65535