3, Tìm x, biết:
a) 2,5 : 7,5 = x : \(\dfrac{3}{5}\)
b) \(2\dfrac{2}{3}\) : x = \(1\dfrac{7}{9}\)
c) \(\dfrac{5}{6}\) : x = 20 : 3
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a) \(\dfrac{27}{2}\cdot7,5+\dfrac{27}{2}\cdot2,5-150\)
\(=\dfrac{27}{2}\cdot\left(7,5+2,5\right)-150\)
\(=\dfrac{27}{2}\cdot10-150\)
\(=135-150\)
\(=-15\)
b) \(3^3\cdot\dfrac{18}{5}-3^3\cdot2\dfrac{2}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\dfrac{18}{5}-3^3\cdot\dfrac{12}{5}-3^3\cdot\dfrac{6}{5}\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{12}{5}-\dfrac{6}{5}\right)\)
\(=3^3\cdot\left(\dfrac{18}{5}-\dfrac{18}{5}\right)\)
\(=3^3\cdot0\)
\(=0\)
a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)
\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)
b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)
\(\Leftrightarrow x+1=\dfrac{50}{9}\)
hay \(x=\dfrac{41}{9}\)
c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
hay \(x\in\left\{8;-8\right\}\)
c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
\(7.9=\left(x-1\right).\left(x+1\right)\)
\(63=x^2-1\)
\(x^2=63+1\)
\(x^2=64\)
\(x^2=8^2\)
\(x=8\)
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, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)
\(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)
\(x\) = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)
\(x\) = \(\dfrac{1}{10}\)
a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x\) = \(\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x\) = \(\dfrac{1}{4}\)
x = \(\dfrac{2}{3}-\dfrac{1}{4}\)
\(x=\dfrac{5}{12}\)
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)
b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)
c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)
d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)
a) x = 3
b) x = \(\dfrac{1}{9}\)
c) x = \(\dfrac{4}{3}\)
d)\(\dfrac{48}{35}\)
a, \(2,5:7,5=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=2,5:7,5\)
=> \(x.7,5=\dfrac{3}{5}.2,5\) => \(x=\dfrac{1,5}{7,5}=\dfrac{1}{5}\)
b, \(2\dfrac{2}{3}:x=1\dfrac{7}{9}\)
=> \(x=2\dfrac{2}{3}:1\dfrac{7}{9}\)
=> \(x=\dfrac{3}{2}\)
c, \(\dfrac{5}{6}:x=20:3\)
=> \(x.20=\dfrac{5}{6}.3\) => \(x=\dfrac{2,5}{20}=\dfrac{1}{8}\)
3, Tìm x, biết:
a) 2,5 : 7,5 = x : \(\dfrac{3}{5}\)
=> \(x:\dfrac{3}{5}=\dfrac{1}{3}\)
=> \(x=\dfrac{1}{3}.\dfrac{3}{5}=>x=\dfrac{1}{5}\)
b) \(2\dfrac{2}{3}\) : x = \(1\dfrac{7}{9}\)
=> \(\dfrac{8}{3}:x=\dfrac{16}{9}\)
=> \(x=\dfrac{8}{3}:\dfrac{16}{9}=>x=\dfrac{3}{2}\)
c) \(\dfrac{5}{6}\) : x = 20 : 3
=> \(x=\dfrac{5}{6}:\dfrac{20}{3}=>x=\dfrac{1}{8}\)