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a) Ta có:
\(\frac{232323}{999999}=\frac{23.10101}{99.10101}=\frac{23}{99}\)
Vì : \(\frac{23}{99}=\frac{23}{99}\\ =>\frac{23}{99}=\frac{232323}{999999}\)
b) \(-\frac{63}{84}=-\frac{3}{4}\\ \frac{65}{-91}=-\frac{5}{7}\)
Vì: \(-\frac{3}{4}< -\frac{5}{7}\\ =>-\frac{63}{84}< \frac{65}{-91}\)
c) Ta có: \(\frac{111}{115}=1-\frac{4}{115}\\ \frac{555}{559}=1-\frac{4}{559}\)
Vì: \(\frac{4}{115}>\frac{4}{559}\\ =>1-\frac{4}{115}< 1-\frac{4}{559}\\ =>\frac{111}{115}< \frac{555}{559}\)
a) \(\dfrac{23}{99}\) và \(\dfrac{232323}{999999}\)
* Giữ nguyên \(\dfrac{23}{99}\).
* Rút gọn \(\dfrac{232323}{999999}=\dfrac{23}{99}\).
Vì \(\dfrac{23}{99}=\dfrac{23}{99}\) nên \(\dfrac{23}{99}=\dfrac{232323}{999999}\)
Vậy \(\dfrac{23}{99}=\dfrac{232323}{999999}\).
b) \(\dfrac{-63}{84}\) và \(\dfrac{65}{-91}\)
\(\circledast\) Rút gọn:
\(\dfrac{-63}{84}=\dfrac{-3}{4}\) ; \(\dfrac{65}{-91}=\dfrac{-5}{7}\)
\(\circledast\) Quy đồng:
Mẫu chung: 28
\(\dfrac{-3}{4}=\dfrac{-3.7}{4\cdot7}=\dfrac{-21}{28}\)
\(\dfrac{-5}{7}=\dfrac{-5\cdot4}{7\cdot4}=\dfrac{-20}{28}\)
Vì \(\dfrac{-21}{28}< \dfrac{-20}{28}\) nên \(\dfrac{-63}{84}< \dfrac{65}{-91}\).
Vậy \(\dfrac{-63}{84}< \dfrac{65}{-91}\).
c) \(\dfrac{111}{115}\) và \(\dfrac{555}{559}\)
\(\dfrac{111}{115}=1-\dfrac{4}{115}\) ; \(\dfrac{555}{559}=1-\dfrac{4}{559}\)
Vì \(\dfrac{4}{115}>\dfrac{4}{559}\)
\(\Rightarrow\) \(1-\dfrac{4}{115}< 1-\dfrac{4}{559}\)
\(\Rightarrow\) \(\dfrac{111}{115}< \dfrac{555}{559}\)
Vậy \(\dfrac{111}{115}< \dfrac{555}{559}\).
tính chất trên gọi là tính chất bắc cầu, ta so sánh hai phân số với một số (phân số) thứ 3.
a: 51/56=1-5/56
61/66=1-5/66
mà -5/56<-5/66
nên 51/56<61/66
b: 41/43<1<172/165
c: \(\dfrac{101}{506}>0>-\dfrac{707}{3534}\)
Câu 1:
a) \(\dfrac{-15}{17}\) và \(\dfrac{-19}{21}\)
Ta có: \(\dfrac{-15}{17}=-1+\dfrac{2}{17}\); \(\dfrac{-19}{21}=-1+\dfrac{2}{21}\)
Vì \(\dfrac{2}{17}>\dfrac{2}{21}\)
Do đó: \(\dfrac{-15}{17}>\dfrac{19}{-23}\)
b) \(\dfrac{-13}{19}\) và \(\dfrac{19}{-23}\)
Ta có: \(\dfrac{19}{23}>\dfrac{19}{25}\); \(\dfrac{13}{19}=1-\dfrac{6}{19}\); \(\dfrac{19}{25}=1-\dfrac{6}{25}\)
mà \(\dfrac{6}{19}>\dfrac{6}{25}\) \(\Rightarrow\dfrac{13}{19}< \dfrac{19}{25}< \dfrac{19}{23}\)
Vì \(\dfrac{13}{19}< \dfrac{19}{23}\Rightarrow\dfrac{-13}{19}>\dfrac{19}{-23}\)
c) \(\dfrac{-24}{35}\) và \(\dfrac{-19}{30}\)
Ta có: \(\dfrac{-24}{35}=-1+\dfrac{19}{35}\);\(\dfrac{-19}{30}=-1+\dfrac{11}{30}\)
Vì \(\dfrac{11}{35}< \dfrac{11}{30}\)
Do đó: \(\dfrac{-24}{35}< \dfrac{-19}{30}\)
d) \(\dfrac{-1941}{1931}\) và \(\dfrac{-2011}{2001}\); \(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)
Vì \(\dfrac{10}{1931}< \dfrac{10}{1001}\)
Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931}\)
Sorry câu d mình viết ngược:
Làm lại:
d) \(\dfrac{-1941}{1931}\) và \(\dfrac{-2011}{2001}\)
Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931};\)
\(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)
Vì \(\dfrac{10}{1931}< \dfrac{10}{1001}\)
Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
=\(\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{1000}{999}\)
=\(\dfrac{1000}{2}\)
=500
Chúc bạn học tốt nha.
\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).....\left(\frac{1}{999}+1\right)\\ < =>\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{1000}{999}\\ < =>\frac{3.4.5....1000}{2.3.4....999}\\ =\frac{1000}{2}=500\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
M > N
N>M