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a) \(\frac{x}{6}=\frac{5}{24}\Rightarrow x=\frac{5}{24}.6=\frac{5}{4}\)
b) \(-\frac{4}{y}=\frac{20}{14}\Rightarrow y=-4:\frac{20}{14}=-\frac{14}{5}\)
c) \(\frac{4}{7}=\frac{12}{x}\Rightarrow x=12:\frac{4}{7}=21\)
d) \(\frac{3}{7}=\frac{y}{21}\Rightarrow y=\frac{3}{7}.21=9\)
\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)) +(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)):\(\frac{7}{5}\)
= \(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))+(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\)).\(\frac{5}{7}\)
=2.\(\frac{5}{7}\).(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{4}{7}\))
=\(\frac{10}{7}\).(\(\frac{21}{42}\)-\(\frac{8}{42}\)+\(\frac{24}{42}\))
=\(\frac{10}{7}\).\(\frac{37}{42}\)=\(\frac{185}{147}\)
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\(\frac{x}{5}=\frac{3}{15}\)suy ra x = 1
\(\frac{x}{7}=\frac{2}{14}\)suy ra x = 1
\(\frac{\left(x+1\right)}{3}=\frac{1}{2}\)suy ra x = \(\frac{1}{2}\)
\(\frac{\left(x-5\right)}{6}=\frac{1}{3}\)suy ra x = 7
\(\frac{\left(x:3\right)}{7}=2\left(\frac{1}{7}\right)\)
\(\frac{\left(x:3\right)}{7}=\frac{15}{7}\)suy ra x = 45
a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)
\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)
hay \(x=-\dfrac{41}{12}\)
Vậy: \(x=-\dfrac{41}{12}\)
a) Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\)
=> (3 + x).5 = 3(5 + y)
=> 15 + 5x = 15 + 3y
=> 5x = 3y
=> x = 3/5y
Mà x + y = 16
hay 3/5y + y = 16
=> (3/5 + 1).y = 16
=> 8/5.y = 16
=> y = 16 : 8/5
=> y = 10
=> x = 16 - 10 = 6
Vậy x = 6; y = 10
b) Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\)
=> (x - 7).6 = 7.(y - 6)
=> 6x - 42 = 7y - 42
=> 6x = 7y
=> x = 7/6y
Mà x - y = -4
hay 7/6y - y = -4
=> 1/6y = -4
=> y = -4 : 1/6
=> y = -24
=> x = -4 - 24 = -28
Vậy x = -28; y = -24
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(\frac{42}{54}=\frac{7}{9}=\frac{7}{x}=\frac{21}{27}=\frac{y}{27}\Rightarrow x=9;y=21.\)
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