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a)\(12< 13;49>47\)
\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)
b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)
c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)
d)
\(1-\dfrac{67}{77}=\dfrac{10}{77}\)
\(1-\dfrac{73}{83}=\dfrac{10}{83}\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)
\(1-\dfrac{123}{128}=\dfrac{5}{128}\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)
b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)
\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)
c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)
d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)
\(\dfrac{73}{83}+\dfrac{10}{83}=1\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)
\(\dfrac{123}{128}+\dfrac{5}{128}=1\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(4)\)
\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)
\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)
\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)
\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)
\(\Leftrightarrow50\left(2x-2\right)=-17.10\)
\(100x-100=-170\)
\(100x=-170+100=-70\)
\(x=-70:100=\dfrac{-7}{10}\)
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\left(x+1\right)\left(x-1\right)5.7\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(x^2-1=35\)
\(x^2=36\)
\(\Leftrightarrow x=\left\{\pm6\right\}\)
bạn có thể giải đc các bài còn lại k ? K phải mk ép bạn đâu nhưng nếu bạn lm đc thì giúp mk nha
e, D = 512+1 /513+ 1 < 1 => 512+1/ 513+1 < 512+1+4/ 513+1+4
= 512+5/ 513+5 = 5. (511+1) / 5. (512+1) = 511+1 / 512+1= E
Vậy D < E
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)