Đáp án:
B2:
\(\dfrac{x}{y} = \dfrac{{29}}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.DK:x \ne \left\{ {\dfrac{1}{2};\dfrac{2}{3}} \right\}\\
\dfrac{3}{{1 - 2x}} = - \dfrac{5}{{3x - 2}}\\
\to 9x - 6 = - 5x + 10\\
\to 14x = 16\\
\to x = \dfrac{8}{7}\left( {TM} \right)\\
b.\left\{ \begin{array}{l}
\dfrac{{ - x}}{3} = \dfrac{y}{5}\\
xy = - \dfrac{5}{{27}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{{5x}}{3}\\
x.\left( { - \dfrac{{5x}}{3}} \right) = - \dfrac{5}{{27}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{{x^2}}}{3} = \dfrac{1}{{27}}\\
y = - \dfrac{{5x}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} = \dfrac{3}{{27}}\\
y = - \dfrac{{5x}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \sqrt {\dfrac{3}{{27}}} = \dfrac{1}{3}\\
x = - \sqrt {\dfrac{3}{{27}}} = - \dfrac{1}{3}
\end{array} \right. \to \left[ \begin{array}{l}
y = - \dfrac{5}{9}\\
y = \dfrac{5}{9}
\end{array} \right.\\
B2:DK:x \ne - 3y\\
\dfrac{{5x - 2y}}{{x + 3y}} = \dfrac{7}{4}\\
\to 10x - 8y = 7x + 21y\\
\to 3x = 29y\\
\to \dfrac{x}{y} = \dfrac{{29}}{3}
\end{array}\)
