Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52+-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
Chúc bạn học tốt!
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
c) \(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-22}{55}-\dfrac{-15}{55}=\dfrac{-7}{55}\)
d) \(\dfrac{-34}{37}.\dfrac{74}{-85}=\dfrac{4}{5}\)
e) \(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{-5}{9}.\dfrac{18}{-7}=\dfrac{10}{7}\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a: \(A=\left[6\cdot\dfrac{1}{27}+3\cdot\dfrac{1}{3}+1\right]:\dfrac{-4}{3}\)
\(=\left(\dfrac{2}{9}+2\right)\cdot\dfrac{-3}{4}\)
\(=\dfrac{20}{9}\cdot\dfrac{-3}{4}=\dfrac{-60}{36}=\dfrac{-5}{3}\)
b: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)
a/ \(5\dfrac{5}{47}+\dfrac{37}{53}+2,7-\dfrac{5}{47}+\dfrac{6}{53}\)
= \(\dfrac{240}{47}-\dfrac{5}{47}+\dfrac{37}{53}+\dfrac{6}{53}+2,7\)
=\(\left(\dfrac{240}{47}-\dfrac{5}{47}\right)+\left(\dfrac{37}{53}+\dfrac{6}{53}\right)+2,7\)
= 5 + \(\dfrac{43}{53}\) + 2,7 = \(\dfrac{4511}{530}\)
b/ \(42\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)-52\dfrac{1}{6}:\left(-1\dfrac{3}{5}\right)\)
= \(\left(42\dfrac{1}{6}-52\dfrac{1}{6}\right):\left(-1\dfrac{3}{5}\right)\)
= \(-10:\left(-1\dfrac{3}{5}\right)\) =25/4
a, Ta có:
\(\dfrac{-13}{39}=\dfrac{-1}{3}\) và \(-\dfrac{21}{63}=\dfrac{-1}{3}\)
Vì \(\dfrac{-1}{3}=\dfrac{-1}{3}\) nên \(\dfrac{-13}{39}=-\dfrac{21}{63}\)
b, Ta có:
\(\dfrac{1}{234567}>0\) (số hữu tỉ dương) và \(-\dfrac{2}{14}< 0\) (số hữu tỉ âm)
=> \(\dfrac{1}{234567}>-\dfrac{2}{14}\)
c\(\dfrac{1}{2012}>-\dfrac{1}{14}\), Ta có:
\(\dfrac{-39}{65}=\dfrac{-3}{5}\) và \(-\dfrac{21}{35}=\dfrac{-3}{5}\)
mà \(\dfrac{-3}{5}=\dfrac{-3}{5}\) nên \(\dfrac{-39}{65}=-\dfrac{21}{35}\)
d,Ta có:
\(\dfrac{1}{2012}>0\) (số hữu tỉ dương) và \(-\dfrac{1}{14}< 0\) (số hữu tỉ âm)
Vậy suy ra: \(\dfrac{1}{2012}>-\dfrac{1}{14}\)
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
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) \(\Rightarrow\)\(\dfrac{-2525}{4949}< \dfrac{-131313}{373737}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=-\dfrac{7}{156}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52\right)+\left(-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
Chúc bạn học tốt!