Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)
Ta có
9/-81=-9/81
Vì -8/9 < -6/7 < -9/81 < 3/4 < 5/6
=> Thứ tự tăng dần là : -8/9 ; -6/7 ; 9/-81 ; 3/4 ; 5/6
Chúc bn hok tốt!
\(\dfrac{-1}{2}=\dfrac{-9}{18};\dfrac{-5}{9}=\dfrac{-10}{18};\dfrac{-1}{3}=\dfrac{-6}{18}\)
mà -10<-9<-6<0
nên \(-\dfrac{5}{9}< -\dfrac{1}{2}< -\dfrac{1}{3}< 0\)(1)
Ta có: \(\dfrac{5}{12}=\dfrac{15}{36};\dfrac{7}{18}=\dfrac{14}{36};\dfrac{1}{3}=\dfrac{12}{36}\)
mà \(0< \dfrac{12}{36}< \dfrac{14}{36}< \dfrac{15}{36}\)
nên \(0< \dfrac{1}{3}< \dfrac{7}{18}< \dfrac{5}{12}\left(2\right)\)
Từ (1),(2) suy ra \(-\dfrac{5}{9}< -\dfrac{1}{2}< -\dfrac{1}{3}< \dfrac{1}{3}< \dfrac{7}{18}< \dfrac{5}{12}\)
a: \(\dfrac{-7}{12}=\dfrac{-7\cdot5}{12\cdot5}=\dfrac{-35}{60};\dfrac{-1}{-15}=\dfrac{1}{15}=\dfrac{4}{60};\dfrac{-5}{4}=\dfrac{-5\cdot15}{4\cdot15}=-\dfrac{75}{60}\)
\(\dfrac{3}{-5}=\dfrac{-3}{5}=\dfrac{-3\cdot12}{5\cdot12}=-\dfrac{36}{60}\)
mà -75<-36<-35<4
nên \(-\dfrac{75}{60}< -\dfrac{36}{60}< -\dfrac{35}{60}< \dfrac{4}{60}\)
=>\(\dfrac{-5}{4}< \dfrac{3}{-5}< \dfrac{-7}{12}< \dfrac{-1}{-15}\)
b: \(\dfrac{-8}{25}+\dfrac{22}{23}+\dfrac{-17}{25}\)
\(=\left(-\dfrac{8}{25}-\dfrac{17}{25}\right)+\dfrac{22}{23}\)
\(=-1+\dfrac{22}{23}=-\dfrac{1}{23}\)
a) Ta có: \(\dfrac{19}{33}=\dfrac{38}{66};\dfrac{6}{12}=\dfrac{1}{2}=\dfrac{33}{66};\dfrac{13}{22}=\dfrac{39}{66}\)
Mà \(\dfrac{33}{66}< \dfrac{38}{66}< \dfrac{39}{66}\Rightarrow\dfrac{6}{12}< \dfrac{19}{33}< \dfrac{13}{22}\)
Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{6}{12};\dfrac{19}{33};\dfrac{13}{22}\)
b) Ta có:
\(\dfrac{-18}{12}=\dfrac{-3}{2}=\dfrac{-105}{70};\dfrac{-10}{7}=\dfrac{-100}{70};\dfrac{-8}{5}=\dfrac{-112}{70}\)
Mà \(\dfrac{-112}{70}< \dfrac{-105}{70}< \dfrac{-100}{70}\Rightarrow\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)
Vậy các số hữu tỉ được sắp xếp theo thứ tự tăng dần là: \(\dfrac{-8}{5};\dfrac{-18}{12};\dfrac{-10}{7}\)
a. \(\dfrac{19}{33};\dfrac{6}{12};\dfrac{13}{22}\) ( \(MC=132\) )
Quy đồng : \(\dfrac{19}{33}=\dfrac{76}{132}\) ; \(\dfrac{6}{12}=\dfrac{66}{132}\) ; \(\dfrac{13}{22}=\dfrac{78}{132}\)
Vì \(\dfrac{66}{132}< \dfrac{76}{132}< \dfrac{78}{132}\) => \(\dfrac{6}{12}< \dfrac{19}{33}< \dfrac{13}{22}\)
b. \(\dfrac{-18}{12};\dfrac{-10}{7};\dfrac{-8}{5}\) ( \(MC=420\) )
Quy đồng : \(\dfrac{-18}{12}=\dfrac{-630}{420}\) ; \(\dfrac{-10}{7}=\dfrac{-600}{420}\) ; \(\dfrac{-8}{5}=\dfrac{-672}{420}\)
Vì : \(\dfrac{-672}{420}< \dfrac{-630}{420}< \dfrac{-600}{420}\) => \(\dfrac{-8}{5}< \dfrac{-18}{12}< \dfrac{-10}{7}\)
1) \(\dfrac{3}{2}\) : \(\dfrac{9}{4}\) =\(\dfrac{3}{2}\)x\(\dfrac{4}{9}\)=\(\dfrac{12}{18}\)=\(\dfrac{2}{3}\)
2)\(\dfrac{48}{55}\) : \(\dfrac{12}{11}\)= \(\dfrac{48}{55}\) x\(\dfrac{11}{12}\)= \(\)\(\dfrac{528}{660}\)=\(\dfrac{4}{5}\)
3)\(\dfrac{7}{10}\) : \(\dfrac{7}{5}\)=\(\dfrac{7}{10}\) x \(\dfrac{5}{7}\)= \(\dfrac{35}{70}\)=\(\dfrac{1}{2}\)
4)\(\dfrac{6}{7}\) : \(\dfrac{8}{7}\) = \(\dfrac{6}{7}\) x \(\dfrac{7}{8}\) =\(\dfrac{42}{56}\)=\(\dfrac{3}{4}\)
Sắp xếp các thương theo thứ tự tăng dần :\(\dfrac{1}{2}\);\(\dfrac{2}{3}\);\(\dfrac{3}{4}\);\(\dfrac{4}{5}\)
\(\Rightarrow\)\(\dfrac{7}{10}\) : \(\dfrac{7}{5}\) ; \(\dfrac{3}{2}\) :\(\dfrac{9}{4}\) ; \(\dfrac{6}{7}\) : \(\dfrac{8}{7}\) ; \(\dfrac{48}{55}\):\(\dfrac{12}{11}\).
\(\dfrac{-1}{12},\dfrac{-3}{4},\dfrac{2}{9},\dfrac{7}{6}\)
\(-\dfrac{1}{12},-\dfrac{3}{4},\dfrac{2}{9},\dfrac{7}{6}\)