M>N

Tham khảo:

https://hoc247.net/hoi-dap/toan-6/so-sanh-m-101-102-1-101-103-1-va-n-101-103-1-101-104-1--faq225210.html

\(M>N\)

giải giúp mink với

M > N

M=101^102+1/101^103+1

M=101^102+1/101^102*101+1

M=1/101+2

M=1/102

N=101^103+1/101^104+1

N=101^103+1/101^103*101+1

N=1/101+1

N=1/102

Vậy N=M

ban làm đề hsg lớp 6 huyện nông cống đúng k.

ta thấy M=101^102+1/101^103+1=101^103+101/101^104+101<101^103/101^104+101(1)

N=101^103+1/101^104+1=101^103/101^104+1/101^104>101^103/101^104+101(2)

tu (1)va (2)suy ra

m<n

\(N=\frac{101^{103}+1}{101^{104}+1}<\frac{101^{103}+1+100}{101^{104}+1+100}=\frac{101^{103}+101}{101^{104}+101}=\frac{101\left(101^{102}+1\right)}{101\left(101^{103}+1\right)}=\frac{101^{102}+1}{101^{103}+1}\)

=> N < M

đinh đức hùng cả cách làm đi bạn

Nhân M và N với 101 ta đc :

101M = \(\frac{101^{103}+101}{101^{103}+1}=\frac{101^{103}+1+100}{101^{103}+1}=1+\frac{100}{101^{103}+1}\)

101N = \(\frac{101^{104}+101}{101^{104}+1}=\frac{101^{104}+1+100}{101^{104}+1}=1+\frac{100}{101^{104}+1}\)

Vì \(101^{103}+1<101^{104}+1\Rightarrow\frac{100}{101^{103}+1}>\frac{100}{101^{104}+1}\)

=> 101M > 101N => M > N.

k nha bạn

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 \(\frac{100}{101^{104}+1}< \frac{100}{101^{103}+1}\Rightarrow101N< 101M\Rightarrow N< M\)

có một số khi nhân số bé lên 10 lần thì số đó là

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

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

=>

M>N K CHO MK NHAAAAAAAAAAAAAAAAAAAAAAA