\(\frac{35}{x}=\frac{15}{12}\Leftrightarrow x=35.12:15=28\)
\(-\frac{2,6}{x}=-\frac{12}{42}\Leftrightarrow\frac{-2,6}{x}=-\frac{2}{7}\Leftrightarrow x=-2,6.7:\left(-2\right)=\frac{91}{10}\)
\(\frac{125}{10}=\frac{x}{45}\Leftrightarrow x=125.45:10=562,5\)
\(\frac{x^2}{6}=\frac{24}{25}\Leftrightarrow x^2=6.24:25=\frac{144}{25}\Leftrightarrow x=\sqrt{\frac{144}{25}}\Leftrightarrow x=\frac{12}{5}\)
1; \(\dfrac{35}{x}\) = \(\dfrac{15}{12}\) (\(x\ne\) 0)
⇒ \(x\) = 35 : \(\dfrac{15}{12}\)
⇒ \(x\) = 28
Vậy \(x=28\)
2; - \(\dfrac{2,6}{x}\) = \(\dfrac{-12}{42}\) (đk \(x\) ≠ 0)
\(x\) = - 2,6 : (\(\dfrac{-12}{42}\))
\(x\) = - \(\dfrac{13}{5}\) : \(\dfrac{-2}{7}\)
\(x\) = - \(\dfrac{13}{5}\) x \(\dfrac{7}{-2}\)
\(x\) = \(\dfrac{91}{10}\)
Vậy \(x\) = \(\dfrac{91}{10}\)
3; \(\dfrac{125}{10}\) = \(\dfrac{x}{45}\)
12,5 = \(\dfrac{x}{45}\)
\(x\) = 12,5 x 45
\(x\) = 562,5
Vậy \(x\) = 562,5
4; \(\dfrac{x^2}{6}\) = \(\dfrac{24}{25}\)
\(x^2\) = 6 x \(\dfrac{24}{25}\)
\(x^2\) = \(\dfrac{144}{25}\)
\(\left[{}\begin{matrix}x=-\dfrac{12}{5}\\x=\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{12}{5}\); \(\dfrac{12}{5}\)}