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5 tháng 9 2021
\(a,2003\cdot2005=\left(2004-1\right)\left(2004+1\right)=2004^2-1< 2004^2\)
\(b,7^{16}-1\\ =\left(7^8-1\right)\left(7^8+1\right)=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =\left(7-1\right)\left(7+1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\\ =48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)>8\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
KK
5 tháng 9 2021
a. Dựa vào tính chất thừa và thiếu, suy ra: 2003 . 2005 = 20042
ND
1
VT
1
MA
20 tháng 9 2016
\(D=8\left(7^8+1\right)\left(7^4+1\right)\left(7^2-1\right)\)
\(D=\frac{4}{25}\left(7^2+1\right)\left(7^2-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^8-1\right)\left(7^8+1\right)\)
\(D=\frac{4}{25}\left(7^{16}-1\right)\)
Vì: \(\frac{4}{25}\left(7^{16}-1\right)< 7^{16}-1\Rightarrow D< C\)
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 \(N=\frac{7^{32}-1}{3}\)
10 tháng 11 2017
\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)
\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)
\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)
KL
1
HH
26 tháng 6 2018
Giải:
a) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=2^{32}-1\)
\(\Leftrightarrow M=\dfrac{2^{32}-1}{3}\)
Vậy ...
b) \(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^{16}-1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=7^{32}-1\)
\(\Leftrightarrow N=\dfrac{7^{32}-1}{3}\)
Vậy ...
28 tháng 6 2018
Bài 1 :
a ) Ta có :
\(\left(x+y\right)^2=x^2+y^2+2xy=20+16=36\)
b ) Ta có :
\(x^2+y^2=\left(x+y\right)^2-2xy=64-30=34\)
Ta có: \(8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)=\frac{1}{6}.48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^8-1\right)\left(7^8+1\right)\)
\(=\frac{1}{6}\left(7^{16}-1\right)\)
Vì \(7^{16}-1>\frac{1}{6}\left(7^{16}-1\right)\) nên \(7^{16}-1>8\left(7^8+1\right)\left(7^4+1\right)\left(7^2+1\right)\)