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14 tháng 9 2021
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4-3}{156}=-\dfrac{7}{156}\)
HD
3 tháng 9 2017
\(\dfrac{\dfrac{1}{6}-\dfrac{1}{39}+\dfrac{1}{51}}{\dfrac{1}{8}-\dfrac{1}{52}+\dfrac{1}{68}}\)
\(\dfrac{11}{\dfrac{78}{\dfrac{11}{104}+\dfrac{1}{68}}}+\dfrac{1}{51}\)
\(\dfrac{71}{\dfrac{442}{\dfrac{213}{1768}}}\)\(\Rightarrow\dfrac{4}{3}\)
16 tháng 6 2021
Bài 1:
\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)
\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\)
\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\)
Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\)
\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\)
\(\dfrac{-18}{31}\)
\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\)
Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)
16 tháng 6 2021
Bài 2:
a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\)
b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)
HD
17 tháng 9 2017
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52+-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
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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: \(B=\dfrac{\dfrac{1}{3}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}{-\dfrac{1}{4}\left(\dfrac{1}{13}-\dfrac{1}{2}-\dfrac{1}{17}\right)}:\dfrac{11}{6}\)
\(=\dfrac{-1}{3}:\dfrac{1}{4}\cdot\dfrac{6}{11}=\dfrac{-4}{3}\cdot\dfrac{6}{11}=\dfrac{-24}{33}=\dfrac{-8}{11}\)
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) \(\dfrac{-5}{9}:\dfrac{-7}{18}=\dfrac{-5}{9}.\dfrac{18}{-7}=\dfrac{10}{7}\)
T
26 tháng 9 2017
\(P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}\\ \)
\(\Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
\(\Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\\ \Rightarrow P=1\)
9 tháng 8 2023
a: =>x*7/4+3/2=-4/5
=>x*7/4=-4/5-3/2=-8/10-15/10=-23/10
=>x=-23/10:7/4=-23/10*4/7=-92/70=-46/35
b: =>x*9/20=1/7+1/8=15/56
=>x=15/56:9/20=15/56*20/9=25/42
c: |x|=3,5
=>x=3,5 hoặc x=-3,5
d: |x|=-2,7
=>x thuộc rỗng
e: =>|x-1|=3-0,73=2,27
=>x-1=2,27 hoặc x-1=-2,27
=>x=-1,27 hoặc x=3,27
f: \(\Leftrightarrow7\cdot11x+11=0\)
=>77x=-11
=>x=-1/7
l: =>|x+3/4|=-2+5=3
=>x+3/4=3 hoặc x+3/4=-3
=>x=-15/4 hoặc x=9/4
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) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=-\dfrac{7}{156}\)
\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-52}{2028}+\dfrac{-39}{2028}=\dfrac{\left(-52\right)+\left(-39\right)}{2028}=\dfrac{-91}{2028}=\dfrac{-7}{156}\)
Chúc bạn học tốt!