so sánh hai số M và N biết:
M=2^2016 và N=(1+2)(2^2 + 1)(2^4 + 1)......(2^2008 + 1)
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DN
2
Những câu hỏi liên quan
5 tháng 9 2021
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=2^{2016}-1< 2^{2016}=M\)
DN
2
5 tháng 9 2021
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
5 tháng 9 2021
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
NV
0
24 tháng 4 2021
Giải:
a)Ta có:
C=1957/2007=1957+50-50/2007
=2007-50/2007
=2007/2007-50/2007
=1-50/2007
D=1935/1985=1935+50-50/1985
=1985-50/1985
=1985/1985-50/1985
=1-50/1985
Vì 50/2007<50/1985 nên -50/2007>-50/1985
⇒C>D
b)Ta có:
A=20162016+2/20162016-1
A=20162016-1+3/20162016-1
A=20162016-1/20162016-1+3/20162016-1
A=1+3/20162016-1
Tương tự: B=20162016/20162016-3
B=1+3/20162016-3
Vì 20162016-1>20162016-3 nên 3/20162016-1<3/20162016-3
⇒A<B
Chúc bạn học tốt!
25 tháng 4 2021
Làm tiếp:
c)Ta có:
M=102018+1/102019+1
10M=10.(102018+1)/202019+1
10M=102019+10/102019+1
10M=102019+1+9/102019+1
10M=102019+1/102019+1 + 9/102019+1
10M=1+9/102019+1
Tương tự:
N=102019+1/102020+1
10N=1+9/102020+1
Vì 9/102019+1>9/102020+1 nên 10M>10N
⇒M>N
Chúc bạn học tốt!
IL
1
5 tháng 8 2018
Câu 2: \(x^2-5x+1=0\Leftrightarrow x^2-2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+1=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{21}{4}=0\Leftrightarrow x-\frac{5}{2}=\pm\frac{\sqrt{21}}{2}\)\(\Leftrightarrow x=\pm\frac{\sqrt{21}+5}{2}\)
Thay vào biểu thức đó:
\(\frac{x^2+1}{x^2}=1+\frac{1}{x^2}=1+\frac{1}{\frac{\left(\sqrt{21}+5\right)^2}{4}}\)
\(=1+\frac{1}{\frac{21+10\sqrt{21}+25}{4}}=1+\frac{4}{46+10\sqrt{21}}=\frac{50+10\sqrt{21}}{46+10\sqrt{21}}\)
\(=\frac{25+5\sqrt{10}}{23+5\sqrt{10}}\). ĐS...
Ta có:
\(N=\left(1+2\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=\left(2^8-1\right)...\left(2^{2008}+1\right)\)
\(\Leftrightarrow N=2^{4016}-1>2^{2016}=M\)
Ta có:
N=(1+2)(2−1)(22+1)(24+1)...(22008+1)N=(1+2)(2−1)(22+1)(24+1)...(22008+1)
⇔N=(22−1)(22+1)(24+1)...(22008+1)⇔N=(22−1)(22+1)(24+1)...(22008+1)
⇔N=(24−1)(24+1)...(22008+1)⇔N=(24−1)(24+1)...(22008+1)
⇔N=(28−1)...(22008+1)⇔N=(28−1)...(22008+1)
⇔N=24016−1>22016=M