\((3+1^{ })^2\div(\dfrac{-1}{4})=\dfrac{-49}{4}\) giúp mình với nha!!! cần gấp lắm

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16 tháng 5 2017

Mik không hiểu câu hỏi cho lắm bn có thể giải thích giúp mik đc hk?????????

16 tháng 5 2017

Tìm x mà

13 tháng 8 2017

a)\(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\)

\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\)

\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)

\(\dfrac{2}{3}x=\dfrac{25}{12}\)

\(x=\dfrac{25}{12}:\dfrac{2}{3}\)

=>\(x=\dfrac{25}{8}\)

13 tháng 8 2017

a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)

\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)

\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)

\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)

\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)

\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)

\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)

\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)

\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)

d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)

\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\left(\dfrac{-2}{3}\right)^3\)

\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{1}{4}x=2\)

\(\Rightarrow x=2:\dfrac{1}{4}\)

\(\Rightarrow x=2.4=8\)

25 tháng 8 2017

a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)

x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)

x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)

x=\(\dfrac{63}{142}\)

25 tháng 8 2017

a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)

\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)

\(x.\dfrac{71}{42}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:\dfrac{71}{42}\)

\(x=\dfrac{63}{142}\)

b\(\dfrac{5}{4}-\dfrac{3}{5}:x=1\dfrac{1}{3}\)

\(\dfrac{3}{5}:x=\dfrac{5}{4}-1\dfrac{1}{3}\)

\(\dfrac{3}{5}:x=\dfrac{-1}{12}\)

\(x=\dfrac{3}{5}:\dfrac{-1}{12}\)

\(x=\dfrac{-36}{5}\)

c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)

\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)

\(\dfrac{4}{7}x-\dfrac{1}{3}=\dfrac{1}{7}\)

\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)

\(\dfrac{4}{7}x=\dfrac{10}{21}\)

\(x=\dfrac{10}{21}:\dfrac{4}{7}\)

\(x=\dfrac{5}{6}\)

d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)

\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)

\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)

\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)

\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)

\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)

\(x=\dfrac{27}{110}\)

có sai sót j xin bn thông cảm !

18 tháng 4 2017

b1:

5:7 được viết : 5/7

1.7:3.12 được viết;85/156

1.5:3.4 được viết: 4/15

-3/1/14:5 đổi -3/1/4= -13/4; đươc viết : -13/20

tick cho mk nhé

15 tháng 3 2017

a)\(\left(5,75\right):x=\dfrac{14}{23}\)

\(\Rightarrow\dfrac{23}{4}:x=\dfrac{14}{23}\)

\(\Rightarrow x=\dfrac{529}{56}\)

b)\(\left(\dfrac{2x}{5}-1\right)\left(-5\right)=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{2x}{5}-1=\dfrac{-1}{20}\)

\(\Rightarrow\dfrac{2x}{5}=\dfrac{19}{20}\)

\(\Rightarrow2x=\dfrac{19}{4}\)

\(\Rightarrow x=\dfrac{19}{8}\)

c)\(\dfrac{x-5}{12,1}=\dfrac{10}{x-5}\)

\(\Rightarrow\left(x-5\right)\left(x-5\right)=10.12,1\)

\(\Rightarrow\left(x-5\right)^2=121\)

\(\Rightarrow\left[{}\begin{matrix}x-5=11\\x-5=-11\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=16\\x=-6\end{matrix}\right.\)

d)

\(2\dfrac{1}{4}x-9\dfrac{1}{4}=20\)

\(\Rightarrow\dfrac{9}{4}x-\dfrac{37}{4}=20\)

\(\Rightarrow\dfrac{9}{4}x=\dfrac{117}{4}\)

\(\Rightarrow x=13\)

16 tháng 3 2017

Hình như câu b c sai sai ớ

8 tháng 5 2017

Ta có :

\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)

\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)

\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)

\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)

\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)

\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)

31 tháng 5 2017

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{1.3}\)

\(...\)

\(\dfrac{1}{100^2}>\dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{100}=\dfrac{99}{100}\\ \dfrac{99}{100}< \dfrac{1}{4}\\ \Rightarrowđpcm\)

a) \(\dfrac{1}{3}x-\dfrac{1}{2}=\dfrac{3}{4}x+\dfrac{1}{15}\)

\(\Rightarrow\dfrac{1}{3}x-\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{1}{15}\)

\(\Rightarrow\dfrac{4}{12}x-\dfrac{9}{12}x=\dfrac{15}{30}+\dfrac{2}{30}\)

\(\Rightarrow\dfrac{-5}{12}x=\dfrac{17}{30}\)

\(\Rightarrow x=\dfrac{-102}{75}\)

\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\dfrac{64}{729}\)

\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)

\(\Rightarrow x=\dfrac{2}{3}\)

16 tháng 3 2018

\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)

\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)

\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)

\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)