Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{4}{x} = \dfrac{x}{9}\,\,\,\,\,\,\,\,\,\,\left( {x \ne 0} \right)\\
\Leftrightarrow 4.9 = x.x\\
\Leftrightarrow {x^2} = 36\\
\Leftrightarrow {x^2} = {6^2}\\
\Leftrightarrow x = \pm 6\\
b,\\
\dfrac{{2x}}{3} = \dfrac{{\dfrac{3}{5}}}{{\dfrac{1}{4}}}\\
\Leftrightarrow 2x.\dfrac{1}{4} = \dfrac{3}{5}.3\\
\Leftrightarrow \dfrac{1}{2}x = \dfrac{9}{5}\\
\Leftrightarrow x = \dfrac{9}{5}:\dfrac{1}{2}\\
\Leftrightarrow x = \dfrac{{18}}{5}\\
c,\\
\dfrac{{x + 1}}{2} = \dfrac{{13}}{5}\\
\Leftrightarrow 5.\left( {x + 1} \right) = 13.2\\
\Leftrightarrow 5x + 5 = 26\\
\Leftrightarrow 5x = 21\\
\Leftrightarrow x = \dfrac{{21}}{5}\\
d,\\
\dfrac{4}{{2x - 1}} = \dfrac{3}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne \dfrac{1}{2}} \right)\\
\Leftrightarrow 4.5 = 3.\left( {2x - 1} \right)\\
\Leftrightarrow 20 = 6x - 3\\
\Leftrightarrow 6x = 23\\
\Leftrightarrow x = \dfrac{{23}}{6}\\
e,\\
\dfrac{{{y^2}}}{4} = \dfrac{{16}}{y}\\
\Leftrightarrow {y^2}.y = 16.4\\
\Leftrightarrow {y^3} = {4^2}.4\\
\Leftrightarrow {y^3} = {4^3}\\
\Leftrightarrow y = 4
\end{array}\)
