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\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
a: \(\dfrac{119}{117}=1+\dfrac{2}{117}\)
\(\dfrac{117}{115}=1+\dfrac{2}{115}\)
mà 2/117<2/115
nên \(\dfrac{119}{117}< \dfrac{117}{115}\)
hay \(-\dfrac{119}{117}>-\dfrac{117}{115}\)
b: \(\dfrac{-22}{35}=\dfrac{-22\cdot177}{35\cdot177}=-\dfrac{3894}{6195}\)
\(\dfrac{-103}{177}=\dfrac{-103\cdot35}{177\cdot35}=\dfrac{-3605}{6195}\)
mà -3894<-3605
nên -22/35<-103/177
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
\(\dfrac{121212}{131313}=\dfrac{121212:10101}{131313:10101}=\dfrac{12}{13}\)
a. \(\dfrac{-18}{91}\) và \(\dfrac{-23}{114}\) ( mẫu chung : 10374 )
Quy đồng : \(\dfrac{-18}{91}=\dfrac{-2052}{10374}\) ; \(\dfrac{-23}{114}=\dfrac{-2093}{10374}\)
Vì \(\dfrac{-2052}{10374}>\dfrac{-2093}{10374}\Rightarrow\dfrac{-18}{91}>\dfrac{-23}{114}\)
Vậy...
b. \(\dfrac{-22}{35}\) và \(\dfrac{-103}{177}\) ( MC = 6195 )
Quy đồng : \(\dfrac{-22}{35}=\dfrac{-3894}{6195};\dfrac{-103}{177}=\dfrac{-3605}{6195}\)
Vì \(\dfrac{-3894}{6195}< \dfrac{-3605}{6195}\Rightarrow\dfrac{-22}{35}< \dfrac{-103}{177}\)
Vậy...
c. \(\dfrac{-22}{45}\) và \(\dfrac{-17}{33}\)(MC=495)
Quy đồng : \(\dfrac{-22}{45}=\dfrac{-242}{495};\dfrac{-17}{33}=\dfrac{-255}{495}\)
Vì \(\dfrac{-242}{495}>\dfrac{-255}{495}\Rightarrow\dfrac{-22}{45}>\dfrac{-17}{33}\)
Vậy
\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)
\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)
\(\dfrac{-22}{45}=-\dfrac{2266}{45.103}>-\dfrac{2295}{45.103}=-\dfrac{51}{103}\)
Ta có: \(\dfrac{-22}{45}=\dfrac{-22\cdot103}{45\cdot103}=\dfrac{-2266}{4635}\)
\(\dfrac{-51}{103}=\dfrac{-51\cdot45}{103\cdot45}=\dfrac{-2295}{4635}\)
mà -2266>-2295
nên \(\dfrac{-22}{45}>\dfrac{-51}{103}\)
b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)
\(=\dfrac{13}{276}\)
\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)
\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)
a)Vì \(\dfrac{21}{52}>0\); \(\dfrac{-213}{523}< 0\)
\(\Rightarrow\)\(\dfrac{21}{52}>\dfrac{-213}{523}\)
b)Ta có :\(\dfrac{22}{35}< 1\) ;\(\dfrac{103}{17}>1\)
\(\Rightarrow\)\(\dfrac{22}{35}< \dfrac{103}{17}\)
c)Ta có :\(\dfrac{-2525}{4949}=\dfrac{-25}{49}\) ;\(\dfrac{-131313}{373737}=\dfrac{-13}{37}\)
Lại có :\(\dfrac{25}{49}>\dfrac{25}{50}=\dfrac{1}{2}\) hay \(\dfrac{-25}{49}< \dfrac{-25}{50}=\dfrac{-1}{2}\)(1)
và \(\dfrac{13}{37}< \dfrac{13}{26}=\dfrac{1}{2}\) hay \(\dfrac{-13}{37}>\dfrac{-13}{26}=\dfrac{-1}{2}\) (2)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{-2525}{4949}< \dfrac{-131313}{373737}\)