Các bạn giúp mk với, còn nốt mấy câu so sánh này nữa thôi, ai nhanh mk k cho

a) 

\(\frac{-17}{243}< 0\) 

\(\frac{1}{1965}>0\)    

\(\frac{-17}{243}< \frac{1}{1965}\)  

b, 

\(\frac{23}{-15}< 0\)    

\(\frac{-17}{-49}>0\)   

\(\frac{23}{-15}< \frac{-17}{-49}\)     

c, 

\(\frac{-2004}{2005}=-1+\frac{1}{2005}\)   

\(\frac{-2005}{2006}=-1+\frac{1}{2006}\)   

Vì \(\frac{1}{2005}>\frac{1}{2006}\)  

Nên \(-1+\frac{1}{2005}>-1+\frac{1}{2006}\)   

Vậy \(\frac{-2004}{2005}>\frac{-2005}{2006}\)

a) Ta có : \(\frac{-17}{243}< 0< \frac{1}{1965}\)

\(\Rightarrow\frac{-17}{243}< \frac{1}{1965}\)

b) Ta có : \(\frac{-17}{-49}=\frac{17}{49}\)

\(\Rightarrow\frac{23}{-15}< 0< \frac{17}{49}\)

\(\Rightarrow\frac{23}{-15}< \frac{17}{49}\)

\(\Rightarrow\frac{23}{-15}< \frac{-17}{-49}\)

c) Ta có : \(\frac{2004}{2005}=\frac{2005}{2005}-\frac{1}{2005}=1-\frac{1}{2005}\)

                \(\frac{2005}{2006}=\frac{2006}{2006}-\frac{1}{2006}=1-\frac{1}{2006}\)

mà \(\frac{1}{2005}>\frac{1}{2006}\)

\(\Rightarrow1-\frac{1}{2005}< 1-\frac{1}{2006}\)

\(\Rightarrow\frac{2004}{2005}< \frac{2005}{2006}\)

\(\Rightarrow\frac{-2004}{2005}>\frac{-2005}{2006}\)

Bài 1

=1093/2187

Bai 2

số nhỏ nhất trong các số trên là:2007/2008

Bai 3

 Ta co :111111/151515=11/15 & 11032/15030=11/15

vì 11/15=11/15 nên 111111/151515=11022=15030

bài 22

111111/151515=11022/15030

bài 15

2004/2005 nhỏ nhất

bài 18

=1093/2187

=1093/2187

câu d thì dễ

Thế giúp vs bn ơi

mk bn lớp mấy

nhân cả C và D với 2005 rồi tách ra so sánh

Ta có : \(2005C=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005D=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì \(\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

=> 2005.C < 2005.D

=> C < D

\(C=\frac{2005^{2005}+1}{2005^{2006}+1}< \frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}=D\)

Vậy \(C< D\)

Ta có : 

\(B=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}< \frac{2004}{2005}+\frac{2005}{2006}=A\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vây \(A>B\)

Chúc bạn học tốt ~

thanks b

Câu này chắc chắn bằng vì hay phân số điều y chan mà 

mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15

Ta có VẾ A

\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)

\(2005\cdot A=1+\frac{2004}{2005^{2006}+1}\)

Ta lại có Vế B :

\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)

\(2005\cdot B=1+\frac{2004}{2005^{2005}+1}\)

Nhìn vào trên , suy ra A < B . 

\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

\(\Rightarrow A< B\)

Vì \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1

Nên \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)

Ta có: \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\)

Nên: \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2004}+1}{2005^{2005}+1}\)

=> A < B

\(2005A=\frac{2005^{2005}+1}{2005^{2006}+1}=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}}\) \(=\frac{2005^{2006}+2014+1}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005^{2004}+1}{2005^{2005}+1}=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(=\frac{2005^{2005}+2004+1}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì \(2005^{2006}+1>2005^{2005}+1\)

Nên \(1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

Hay A < B 

           Vậy A < B 

sửa chỗ \(\frac{2005^{2006}+2014+1}{2005^{2006}+1}\) thành \(\frac{2005^{2006}+2004+1}{2005^{2006}+1}\)nhé