Bài 98 : Giải hệ phương trình :
$ \left\{\begin{matrix} { 27x^3y^3+125=9y^3} \\ { 45x^2y+75x=6y^2 } \end{matrix}\right. $ (1)
$y=0$ không là nghiệm. Khi đó xét $y \ne 0$
$$\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
27{x^3} + \dfrac{{125}}{{{y^3}}} = 9\\
45\dfrac{{{x^2}}}{y} + 75\dfrac{x}{{{y^2}}} = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\left( {3x} \right)^3} + {\left( {\dfrac{5}{y}} \right)^3} = 9\\
3x.\dfrac{5}{y}\left( {3x + \dfrac{5}{y}} \right) = 6
\end{array} \right.\,\,\,\,\,\left( 2 \right)$$
Đặt: $$u = 3x + \dfrac{5}{y},\,\,v = 3x.\dfrac{5}{y} \Rightarrow \left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}
{u^3} - 3uv = 9\\
uv = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u^3} = 27\\
uv = 6
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
u = 3\\
v = 2
\end{array} \right.$$
Khi đó: $$\left\{ \begin{array}{l}
3x + \dfrac{5}{y} = 3\\
3x.\dfrac{5}{y} = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x = 2\\
\dfrac{5}{y} = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
3x = 1\\
\dfrac{5}{y} = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \dfrac{2}{3}\\
y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x = \dfrac{1}{3}\\
y = \dfrac{5}{2}
\end{array} \right.
\end{array} \right.$$
Vậy HPT đã cho có nghiệm $\left( {x;y} \right) = \left( {\dfrac{2}{3};5} \right),\,\,\left( {\dfrac{1}{3};\dfrac{5}{2}} \right)$.