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b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
a)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)
b)
\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)
\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)
\(\dfrac{19}{20}x=1\)
\(x=\dfrac{20}{19}\)
Vậy \(x=\dfrac{20}{19}\)
b) \(\left(x^2-9\right)\left(3-5x\right)=0\)
TH1:
\(x^2-9=0\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
=>\(x\in\left\{3;-3\right\}\)
TH2:
\(3-5x=0\)
\(5x=3\)
\(x=\dfrac{3}{5}\)
Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
a, \(\left(\dfrac{7}{2}-2x\right).\dfrac{10}{3}=\dfrac{22}{3}\Leftrightarrow\dfrac{7}{2}-2x=\dfrac{22}{10}=\dfrac{11}{5}\)
\(\Leftrightarrow2x=\dfrac{13}{10}\Leftrightarrow x=\dfrac{13}{20}\)
b, \(\dfrac{4x}{9}=\dfrac{9}{8}-\dfrac{125}{1000}=1\Leftrightarrow x=\dfrac{9}{4}\)
c, \(-\dfrac{x}{21}=\dfrac{60}{21}\Rightarrow x=-60\)