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Ta có:
\(M=\frac{101^{102}+1}{101^{103}+1}\)
\(101M=\frac{101^{103}+1+100}{101^{103}+1}=1+\frac{100}{101^{103}+1}\)
Ta lại có:
\(N=\frac{101^{103}+1}{101^{104}+1}\)
\(101N=\frac{101^{104}+1+100}{101^{104}+1}=1+\frac{100}{101^{104}+1}\)
Vì \(\frac{100}{101^{104}+1}< \frac{100}{101^{103}+1}\Rightarrow101N< 101M\Rightarrow N< M\)
Ta có: M =\(\frac{101^{102}+1}{101^{103}+1}=\frac{101^{103}+101}{101^{104}+101}=\frac{101^{103}+1+100}{101^{104}+1+100}\)
Mà : N = \(\frac{101^{103}+1}{101^{104}+1}\)< M = \(\frac{101^{103}+1+100}{101^{104}+1+100}\)
\(\Rightarrow N< M\)
Ta có : \(101M=\frac{101\left(101^{102}+1\right)}{101^{103}+1}=\frac{101^{103}+100+1}{101^{103}+1}=1+\frac{100}{101^{103}+1};\)
\(101N=\frac{101\left(101^{103}+1\right)}{101^{104}+1}=\frac{101^{104}+1+100}{101^{104}+1}=1\frac{100}{101^{104}+1}\)
Vì \(\frac{100}{101^{103}+1}>\frac{100}{101^{104}+1}\Rightarrow1+\frac{100}{101^{103}+1}>1+\frac{100}{101^{104}+1}\Rightarrow101M>101N\)
=> M > N
So sánh M và N biết rằng :
\(M=\frac{101^{102}+1}{101^{103}+1}\)
\(N=\frac{101^{103}+1}{101^{104}+1}\)
ta có bổ đề sau .với\(\frac{a}{b}>0\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Rightarrow N=\frac{101^{103}+1}{101^{104}+1}< \frac{101^{103}+1+100}{101^{104}+1+100}\)
mà \(\frac{101^{103}+1+100}{101^{104}+1+100}=\frac{101^{103}+101}{101^{104}+101}\)
\(=\frac{101\left(101^{102+1}\right)}{101\left(101^{103}+1\right)}=\frac{101^{102}+1}{101^{103}+1}=M\)
vậy \(M>N\)
Ta có: \(N=\frac{101^{103}+1}{101^{104}+1}< \frac{101^{103}+1+100}{101^{104}+1+100}\)
Mà: \(\frac{101^{103}+1+100}{101^{104}+1+100}=\frac{101^{103}+101}{101^{104}+101}=\frac{101\left(101^{102}+1\right)}{101\left(101^{103}+1\right)}=\frac{101^{102}+1}{101^{103}+1}=M\)
Ta có: \(N< \frac{101^{103}+1+100}{101^{104}+1+100};\frac{101^{103}+1+100}{101^{104}+1+100}=M\)
=> N<M
=>
Ta có :
\(N=\frac{101^{103}+1}{101^{104}+1}< 1=\frac{101^{103}+1+100}{101^{104}+1+100}=\frac{101^{103}+101}{101^{104}+101}=\frac{101\left(101^{102}+1\right)}{101\left(101^{103}+1\right)}=\frac{101^{102}+1}{101^{103}+1}=M\)
Vậy\(N< M\)
ta có:N<1
=> 101103+1/101104+1 <101103+1+100/101104+1+100
<=> N<101103+101/101104+101
<=> N<101.(101102+1)/101.(101103+1)
<=> N<101102+1/101103+1
hayN<M
Vậy N<M
cô giáo dạy mk cách này đó!nếu bn thấy đúng thì ks cho mk nha!
Nếu a/b<1 thì a+m/b+m > a/b (m thuộc Z )
N =101^103+1/101^104+1 < 101^103 +1+100/101^104+1+100
=101^103+101/101^104+101=101x(101^102+1)/101x(101^103+1)
=101^102+1/101^103+1=M
Vậy M < N
Ta có:\(y=\frac{101^{102}+1}{101^{102}+1}\). \(\Rightarrow\)\(101y=\frac{101\left(101^{102}+1\right)}{101^{103}+1}=\frac{101^{103}+101}{101^{103}+1}=1+\frac{100}{101^{103}+1}\)
\(x=\frac{101^{103}+1}{101^{104}+1}\Rightarrow101x=\frac{101\left(101^{103}+1\right)}{101^{104}+1}=\frac{101^{104}+101}{101^{104}+1}=1+\frac{100}{101^{104}+1}\) Vì \(\frac{100}{101^{103}+1}>\frac{100}{101^{104}+1}\)nên \(1+\frac{100}{101^{^{103}}+1}>1+\frac{100}{101^{104}+1}\)hay 101y>101x. Suy ra y>x
A = \(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)< \(\frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\) =\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
= \(\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)= \(\frac{1}{2^2.3.5^2.7}\)= B
Vậy A < B
\(A