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16 tháng 5 2018
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
16 tháng 5 2018
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
30 tháng 7 2017
b) \(\dfrac{x}{27}=\dfrac{3}{x}\)
\(\Rightarrow x.x=27.3\)
\(\Rightarrow x^2=81\)
\(\Rightarrow x^2=9^2\)
\(\Rightarrow x=9\)
Vậy x=9
Q
11 tháng 4 2017
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
11 tháng 4 2017
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
23 tháng 6 2022
a: \(\dfrac{x+2}{27}=\dfrac{x}{-9}\)
=>x+2=-3x
=>4x=-2
hay x=-1/2
b: \(\dfrac{-7}{x}=\dfrac{21}{34-x}\)
=>-7(34-x)=21x
=>34-x=-3x
=>2x=-34
hay x=-17
c: \(\dfrac{-8}{15}< \dfrac{x}{40}< \dfrac{-7}{15}\)
\(\Leftrightarrow-64< 3x< -56\)
hay \(x\in\left\{-21;-20;-19\right\}\)
d: \(\dfrac{-1}{2}< \dfrac{x}{18}< \dfrac{-1}{3}\)
=>-9<x<-6
hay \(x\in\left\{-8;-7\right\}\)
26 tháng 3 2017
1.
\(\dfrac{30}{100}.x+\dfrac{1}{4}=\dfrac{1}{5}.x-\dfrac{1}{2}\)
\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{4}\)
\(\dfrac{1}{5}.x-\dfrac{3}{10}.x=\dfrac{1}{4}\)
\(\left(\dfrac{1}{5}-\dfrac{3}{10}\right).x=\dfrac{1}{4}\)
\(\dfrac{-1}{10}.x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}:\dfrac{-1}{10}\)
x=\(\dfrac{5}{-2}\)=\(\dfrac{-5}{2}\)
2.
\(\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=\left(0,25-3,5\right).\dfrac{27}{3}\)
\(\dfrac{2}{2}.\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=-3,25.9\)
\(\dfrac{1}{2}.\left(\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{31.33}\right).x=-29,25\)
\(\dfrac{1}{2}.\left(\dfrac{1}{7}-\dfrac{1}{33}\right).x=-29,25\)
\(\dfrac{1}{2}.\dfrac{26}{231}.x=-29,25\)
\(\dfrac{13}{231}.x=-29,25\)
\(x=-29,25:\dfrac{13}{231}\)
\(x=\dfrac{-2079}{4}\)
tick mink nha :)
11 tháng 4 2017
\(\dfrac{48}{25}\cdot\dfrac{27}{55}+2\dfrac{4}{9}\cdot\dfrac{14}{33}\)
\(=\dfrac{1296}{1375}+\dfrac{22}{9}\cdot\dfrac{14}{33}\\ =\dfrac{1296}{1375}+\dfrac{28}{27}\\ =\dfrac{34992}{37125}+\dfrac{38500}{37125}\\ =\dfrac{73492}{37125}\)
\(1\dfrac{19}{22}\cdot\left(\dfrac{47}{77}-\dfrac{16}{15}\right)\\ =\dfrac{41}{22}\cdot\dfrac{-527}{1155}\\ =\dfrac{-21607}{25410}\)
\(\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)-\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{307}{99}+\dfrac{37}{99}-\dfrac{1503}{299}\right)-0\\ =\dfrac{344}{99}-\dfrac{1053}{299}\\ =-\dfrac{107}{2277}\)
11 tháng 4 2017
bài 1:
a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)
\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)
\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)
\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)
\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)
c) Cho mình hỏi x ở đâu vậy ???
d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)
\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)
\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)
f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)
\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)
\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)
\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)
\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)
\(x=\dfrac{-5}{7}\)
bài 2:
Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)
\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)
Vậy \(a=-15;b=81\)
20 tháng 7 2018
a/ \(\dfrac{1}{3}+\dfrac{1}{2}:x=4\)
\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2x}=4\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{11}{3}\)
\(\Leftrightarrow22x=3\)
\(\Leftrightarrow x=\dfrac{3}{22}\)
Vậy ...
b/ \(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
Vậy ...
c/ \(\left(2x+1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=4^2\\\left(2x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=4\\2x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy..
d/ \(\dfrac{x+3}{2}=\dfrac{3}{5}\)
\(\Leftrightarrow5\left(x+3\right)=6\)
\(\Leftrightarrow5x+15=6\)
\(\Leftrightarrow x=-\dfrac{9}{5}\)
Vậy..