Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\frac{29}{60}< \frac{1}{2}< \frac{15}{28}\)
b)\(\frac{13}{30}< \frac{1}{2}< \frac{23}{42}\)
c)\(\frac{13}{36}>\frac{1}{3}>\frac{14}{45}\)
d)\(\frac{1919}{9595}=\frac{1}{5}< \frac{11}{50}=\frac{1111}{5050}\)
chúc các bạn học tôt nhớ k đúng cho mình nha !!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a, Ta có: \(\frac{29}{60}< \frac{30}{60}=\frac{1}{2}=\frac{14}{28}< \frac{15}{28}\) . Vậy \(\frac{29}{60}< \frac{15}{28}\)
b, \(\frac{13}{30}< \frac{15}{30}=\frac{1}{2}=\frac{21}{42}< \frac{23}{42}\).Vậy \(\frac{13}{30}< \frac{23}{42}\)
c, \(\frac{13}{36}>\frac{12}{36}=\frac{1}{3}=\frac{15}{45}>\frac{14}{45}\).Vậy \(\frac{13}{36}>\frac{14}{45}\)
d, \(\frac{1919}{9595}=\frac{1}{5}=\frac{1010}{5050}< \frac{1111}{5050}.\)Vậy \(\frac{1919}{9595}< \frac{1111}{5050}\)
a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)
=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)
=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)
=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)
b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)
=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))
=99+1=100
c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha
\(a,\dfrac{-5}{13}+\dfrac{8}{13}=\dfrac{3}{13}\\ b,\dfrac{5}{31}+\dfrac{-22}{31}=\dfrac{-17}{31}\\ c,\dfrac{-13}{43}+\dfrac{-40}{43}=\dfrac{-53}{43}\\ d,\dfrac{-3}{29}-\dfrac{16}{58}=\dfrac{-11}{29}\\ e,\dfrac{8}{40}-\dfrac{-36}{45}=1\\ f,\dfrac{-8}{18}-\dfrac{-15}{27}=\dfrac{1}{9}\\ g,\left(-2\right)+\dfrac{-5}{8}=\dfrac{-21}{8}\)
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: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
Theo quy ước với mọi phân số lớn hơn 0 thì ta có:
\(\dfrac{a}{b}>0=>\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N;n\ne0\right)\)
Áp dụng với bài trên ta => ĐPCM
CHÚC BẠN HỌC TỐT.......