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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
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
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 \(\dfrac{-1}{-3}\)
\(\dfrac{-27}{463}< 0;\dfrac{-1}{-3}=\dfrac{1}{3}>0\)
\(\Rightarrow\dfrac{-27}{463}< \dfrac{-1}{-3}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
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...
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 :
\(\dfrac{2}{3}\&\dfrac{5}{7}\)
\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)
\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{3}\) Câu 3 : \(\dfrac{8}{13}\&\dfrac{5}{7}\)Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)
\(\Rightarrow\dfrac{8}{13}< \dfrac{5}{7}\)
\(-\dfrac{265}{317}< -\dfrac{83}{317}< -\dfrac{83}{111}\Rightarrow-\dfrac{265}{317}< -\dfrac{83}{111}\)
b)Ta có :
\(\dfrac{2002}{2003}< 1< \dfrac{14}{13}\Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)
c)Ta có :
\(\dfrac{-1}{-3}=\dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< 0< \dfrac{1}{3}\Rightarrow-\dfrac{27}{463}< \dfrac{1}{3}\)
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