a,

Ta có:
\(\dfrac{3}{7}=1-\dfrac{4}{7}\)
\(\dfrac{11}{15}=1-\dfrac{4}{15}\)
So sánh phân số \(\dfrac{4}{7}\) và \(\dfrac{4}{15}\)
Vì \(7< 15\) nên \(\dfrac{1}{7}>\dfrac{1}{15}\)
\(\Rightarrow1-\dfrac{4}{7}< 1-\dfrac{4}{15}\)
Vậy \(\dfrac{3}{7}< \dfrac{11}{15}\)

b)

\(\dfrac{-11}{6}< -1< \dfrac{-8}{9}\) nên \(\dfrac{-11}{6}< \dfrac{-8}{9}\)

c) 

\(\dfrac{305}{25}=\dfrac{305:5}{25:5}=\dfrac{61}{5}\)
Ta có: 
Mẫu số chung 2 phân số: 80
\(\dfrac{297}{16}=\dfrac{297*5}{16*5}=\dfrac{1485}{80}\)
\(\dfrac{61}{5}=\dfrac{61*16}{5*16}=\dfrac{976}{80}\)
Vì \(1485>976\) nên\(\dfrac{1485}{80}>\dfrac{976}{80}\)
Vậy \(\dfrac{297}{16}>\dfrac{305}{25}\)

d,

$\frac{-205}{317}=\frac{-205:-1}{317:-1}=\frac{205}{-317}$
Ta có: 
Mẫu số chung 2 phân số: -35187
\(\dfrac{205}{-317}=\dfrac{205*111}{-317*111}=\dfrac{22755}{-35187}\)
\(\dfrac{-83}{111}=\dfrac{-83*-317}{111*-317}=\dfrac{26311}{-35187}\)
Vì \(22755< 26311\) nên\(\dfrac{22755}{-35187}< \dfrac{26311}{-35187}\)
Vậy \(\dfrac{-205}{317}< \dfrac{-83}{111}\)

Câu d, mình làm sai, cho mình sửa lại:

\(\dfrac{-205}{317}=\dfrac{-22755}{35187}\)

\(\dfrac{-83}{111}=\dfrac{-26311}{35187}\)

Vậy là  \(-22755>-26311\) hay \(\dfrac{-205}{317}>\dfrac{-83}{111}\)

a)-17/23=-171717/232323

b)-265/317<-83/111

c)2002/2003<14/13

d)-27/463<1/3

Ta có

Ta có: \(\dfrac{2002}{2003}< \dfrac{14}{13}\) vì \(\dfrac{2002}{2003}< 1\), \(\dfrac{14}{13}>1\)

ta có

2002/2003<1<14/13

=>2002/2003<14/13

a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)

\(\dfrac{-5}{6}>\dfrac{-91}{104}\)

Sao bn giống BT mình thế ?:)

a)\(\dfrac{2002}{2003}\) và \(\dfrac{14}{13}\)

\(\dfrac{2002}{2003}< 1;\dfrac{14}{13}>1\)

\(\Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)

b)\(\dfrac{-33}{37}\) và \(\dfrac{-34}{35}\)

Với phân số âm ,phân số nào cũng tử mà khác mẫu ,mẫu nào lớn hơn thì lớn hơn

\(\Rightarrow\dfrac{-33}{37}>\dfrac{-33}{35}\)

c)\(\dfrac{-27}{463}\) và \(\dfrac{-1}{-3}\)

\(\dfrac{-27}{463}< 0;\dfrac{-1}{-3}=\dfrac{1}{3}>0\)

\(\Rightarrow\dfrac{-27}{463}< \dfrac{-1}{-3}\)

Giúp mình với

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy x = -1

b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

Vậy...

a. \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

Ta thấy: \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}\) nên biểu thức trong dấu ngoặc thứ hai khác 0. Do đó x + 1 = 0 => x = -1

b. \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

=> x = -2004

Câu 1 : 

\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)

Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)

mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)

\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)

Câu 2 :

\(\dfrac{2}{3}\&\dfrac{5}{7}\)

\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)

Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)

`#3107.101107`

`a)`

Ta có:

\(\dfrac{2727}{3131}=\dfrac{2727\div27}{3131\div31}=\dfrac{27}{31}\)

Vì \(\dfrac{27}{31}=\dfrac{27}{31}\)

\(\Rightarrow\dfrac{27}{31}=\dfrac{2727}{3131}\)

`b)`

Ta có:

\(\dfrac{11}{31}=1-\dfrac{20}{31}=1-\dfrac{200}{310}\)

\(\dfrac{111}{311}=1-\dfrac{200}{311}\)

Vì \(\dfrac{200}{310}>\dfrac{200}{311}\)

\(\Rightarrow1-\dfrac{200}{310}< 1-\dfrac{200}{311}\)

\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}.\)

27/31 = 2727/3131

11/31 bé hơn 111/311