Ta có:

1-1995/1997=1997/1997-1995/1997=1/1997

1-1999/2001=2001/2001-1999/2001=1/2001

Vì 1/1995 > 1/2001 nên 1999/2001<1995/1997

 hãy t cho mk nha mn mk biết ơn mn nhìu lắm

Ta có:

\(1-\frac{1995}{1997}=\frac{2}{1997}>1-\frac{1999}{2001}=\frac{2}{2001}\)

Vậy nên:

\(\frac{1995}{1997}<\frac{1999}{2001}\)

Ta có:

1-1995/1997=2/1997>1-1999/2001

=>1995/1997<1999/2001

k cho mình nha

hi hi câu hỏi này mình có đáp án la 1995/1997<1999/2001

43/171 > 45/169

3/11 > 1999/2001

1997/2001 > 2001/2005

giải thích rõ

Ta có: 1 - 1995/1997 = 2/1997

1 - 1999/2001 = 2/2001

mà 2/1997 > 2/2001 nên 1995/1997 < 1999/2001

Ta có: 1 - 1995 / 1997 = 2/1997

          1- 1999/2001= 2/2001

VÌ 2/1997 > 2/2001 Nên 1995/1997 < 1999/2001 

đề dễ quá

\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))

\(\Leftrightarrow x=-2004\)

\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)

\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)

\(\Leftrightarrow x=-2004\)

Vậy PT có tập nghiệm S = {-2004}

\(\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)

<=>\(\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)

<=>\(\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

Mà \(\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)< 0\)

=> X+2004=0

=>X=-2004

cái cuối là =-4 nhé!

\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4\)

\(\Rightarrow\frac{x+2001}{5}+1+\frac{x+1999}{7}+1+\frac{x+1997}{9}+1+\frac{x+1995}{11}+1=0\)

\(\Rightarrow\frac{x+2006}{5}+\frac{x+2006}{7}+\frac{x+2006}{9}+\frac{x+2006}{11}=0\)

\(\Rightarrow\left(x+2006\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)=0\)

\(\Rightarrow x+2006=0\)vì \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}>0\)

\(\Rightarrow x=-2006\)

\(\frac{1997}{1996}>1;\frac{1998}{1999}< 1\Rightarrow\frac{1997}{1996}>\frac{1998}{1999}\)

Ta thấy : \(\frac{1997}{1996}>1;\frac{1998}{1999}< 1\)

\(\rightarrow\frac{1997}{1996}>\frac{1998}{1999}\)

~ ai tk mk mk tk lại cho nha ~

Ta có:

\(\frac{1997}{1996}>\frac{1996}{1996}=1.\)

Mà \(\frac{1998}{1999}< \frac{1999}{1999}=1.\)

=>\(\frac{1997}{1996}< \frac{1998}{1999}.\)

Vậy.......