xy - x + 2y = 3

=> x(y-1) + 2y - 2 = 3 + 2

=> x(y-1) + 2(y-1) = 5

=> (x+2)(y+1) = 5

=> x + 2 và y + 1 \(\in\)Ư(5) = {-1;5;-5;1}

ta có bảng :

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

giải giúp mink với

M > N

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\(N< M\)

Kết quả là:N<M

Ta có :

\(M=\frac{101^2+1}{101^3+1}\)

\(N=\frac{101^3+1}{101^4+1}< 1=\frac{101^3+1+100}{101^3+1+100}=\frac{101^3+101}{101^3+101}=\)\(\frac{101^2\cdot101+101\cdot1}{101^3\cdot101+101\cdot1}=\frac{101\cdot\left(101^2+1\right)}{101\cdot\left(101^3+1\right)}=\frac{101^2+1}{101^3+1}=M\)

\(\Rightarrow M>N\)

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

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

\(M>N\)