Đặt \(A=\frac{1005}{1006}+\frac{1006}{1007}+\frac{1007}{1008}+\frac{1008}{1005}\) ta có : 

\(A=\frac{1006-1}{1006}+\frac{1007-1}{1007}+\frac{1008-1}{1008}+\frac{1005+3}{1005}\)

\(A=\frac{1006}{1006}-\frac{1}{1006}+\frac{1007}{1007}-\frac{1}{1007}+\frac{1008}{1008}-\frac{1}{1008}+\frac{1005}{1005}+\frac{3}{1005}\)

\(A=1-\frac{1}{1006}+1-\frac{1}{1007}+1-\frac{1}{1008}+1+\frac{3}{1005}\)

\(A=\left(1+1+1+1\right)-\left(\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}-\frac{3}{1005}\right)\)

\(A=4-\left(\frac{1}{1006}+\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1005}-\frac{1}{1005}-\frac{1}{1005}\right)\)

\(A=4-\left[\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)\right]\)

Mà : 

\(\frac{1}{1006}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1006}-\frac{1}{1005}< 0\) \(\left(1\right)\)

\(\frac{1}{1007}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1007}-\frac{1}{1005}< 0\) \(\left(2\right)\)

\(\frac{1}{1008}< \frac{1}{1005}\)\(\Rightarrow\)\(\frac{1}{1008}-\frac{1}{1005}< 0\) \(\left(3\right)\)

Từ (1), (2) và (3) suy ra : 

\(\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)< 0\)

\(\Rightarrow\)\(A=4-\left[\left(\frac{1}{1006}-\frac{1}{1005}\right)+\left(\frac{1}{1007}-\frac{1}{1005}\right)+\left(\frac{1}{1008}-\frac{1}{1005}\right)\right]>4\)

\(\Rightarrow\)\(A>4\) ( điều phải chứng minh ) 

Vậy \(A>4\)

Chúc bạn học tốt ~ 

\(\dfrac{1004}{1005}< \dfrac{1005}{1006}< \dfrac{1006}{1007}< \dfrac{1007}{1008}\)

1004/1005<1005/1006<1006/1007<1007/1008

Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)

\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)

\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)

\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)

mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)

nên x+2013=0

hay x=-2013

Vậy: S={-2013}

\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)

\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)

\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)

Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)

\(\Rightarrow x+2013=0\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

1-

Số tự nhiên lẻ nhỏ nhất là 1

Số tự nhiên lẻ lớn nhất những không lớn hơn 2015 là 2015

Khoảng cách giữa hai số lẻ liên tiếp là 2

=> có: ( 2015-1) ; 2 + 1= 1008

2-

Có : ( 2015-0) ; 1 + 1= 2016 

1)b,d

2)a,b,c,d

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}\)

\(=\frac{2^{1005}.7^{1005}.5^{1006}}{2^{1007}.5^{1004}.7^{1004}}\)

\(=\frac{5^2.7}{2^2}=\frac{25.7}{4}=\frac{175}{4}\)

TRẦN TIỂU HY ƠI, BẠN TRÌNH BÀY RA GIÙM MK NHA. MK KO HIỂU LẮM

\(\frac{14^{1005}.5^{1006}}{2^{1007}.35^{1004}}=\frac{2^{1005}.7^{1005}.5^{1006}}{2^{1007}.5^{1004}.7^{1004}}\)

\(=\frac{5^2.7}{2^2}=\frac{25.7}{4}=\frac{175}{4}\)