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HD
17 tháng 9 2017
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52+-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
Chúc bạn học tốt!
HD
3 tháng 9 2017
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
9 tháng 6 2017
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) \(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{-5}{9}.\dfrac{18}{-7}=\dfrac{10}{7}\)
30 tháng 6 2022
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: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)
22 tháng 7 2017
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 !!!
23 tháng 7 2017
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
OT
17 tháng 7 2017
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
18 tháng 7 2017
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
16 tháng 5 2017
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
28 tháng 6 2017
a, \(\dfrac{5}{6}-\left|2-x\right|=\dfrac{1}{3}\Rightarrow\dfrac{5}{6}-\dfrac{1}{3}=\left|2-x\right|\)
<=> \(\dfrac{1}{2}=\left|2-x\right|\) \(\Leftrightarrow\left[{}\begin{matrix}2-x=\dfrac{1}{2}\\2-x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
==================
Mấy câu sau tương tự thôi
21 tháng 1 2018
a)\(\dfrac{3}{2}hay\dfrac{-3}{2}\)
b)\(\dfrac{13}{20}hay\dfrac{-13}{20}\)
c)\(\dfrac{11}{6}hay\dfrac{-11}{6}\)
d)\(\dfrac{4}{3}hay\dfrac{-4}{3}\)
e)\(\dfrac{1}{5}hay\dfrac{-1}{5}\)
Đây là câu trả lời của mình
Hay có nghĩa là hoặc
FC
9 tháng 11 2018
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: \(\dfrac{1}{2012}>-\dfrac{1}{14}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
\(\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{-7}{156}\)