$$ \left\lbrace\begin{array}{l} \dfrac{1}{x}+ \dfrac{1}{y}=9 \\ \left( \dfrac{1}{\sqrt[3]{x}}+ \dfrac{1}{\sqrt[3]{y}}\right)\left( \dfrac{1}{\sqrt[3]{x}}+1\right)\left( \dfrac{1}{\sqrt[3]{y}}+1\right)=18 \end{array}\right.$$

Đề bài

\begin{equation} \label{eq:69.I} \left\lbrace\begin{array}{l}  \dfrac{1}{x}+ \dfrac{1}{y}=9 \\ \left( \dfrac{1}{\sqrt[3]{x}}+ \dfrac{1}{\sqrt[3]{y}}\right)\left( \dfrac{1}{\sqrt[3]{x}}+1\right)\left( \dfrac{1}{\sqrt[3]{y}}+1\right)=18 \end{array}\right.\end{equation}

Lời giải

Đặt $ \dfrac{1}{\sqrt[3]{x}}=a$, $ \dfrac{1}{\sqrt[3]{y}}=b$
\begin{align} \eqref{eq:69.I} &\Rightarrow \left\lbrace\begin{array}{l} a^3+b^3=9 \\ \left(a+b\right)\left(a+1\right)\left(b+1\right)=18 \end{array}\right. \nonumber\\ \label{eq:69.II} &\Leftrightarrow \left\lbrace\begin{array}{l} (a+b)^3-3ab(a+b)=9 \\ \left(a+b\right)\left(ab+a+b+1\right)=18 \end{array}\right. \end{align}

Lại đặt $S=a+b$, $P=ab$
\begin{align} \eqref{eq:69.II} &\Rightarrow \left\lbrace\begin{array}{l} S^3-3SP=9 \\ S(S+P+1)=18 \end{array}\right. \nonumber\\ &\Leftrightarrow \left\lbrace\begin{array}{l} S^3-3SP=9 \\ S^3+3S^2+3S-63=0 \end{array}\right. \nonumber \\ &\Leftrightarrow \left\lbrace\begin{array}{l} S^3-3SP=9 \\ (S-3)\left[ (S+3)^2+12\right] =0 \end{array}\right. \nonumber \\ &\Leftrightarrow \left\lbrace\begin{array}{l} P=2 \\ S=3 \end{array}\right. \nonumber \end{align}

Áp dụng định lý $Viete$ đảo ta có $a,b$ là nghiệm của phương trình $$X^2-3X+2=0 \Leftrightarrow \left[\begin{array}{l} X=2 \\ X=1\end{array}\right. \Leftrightarrow \left[\begin{array}{l} \left\lbrace\begin{array}{l} a=2 \\ b=1\end{array}\right. \\ \left\lbrace\begin{array}{l} a=1 \\ b=2\end{array}\right.\end{array}\right.\Leftrightarrow \left[\begin{array}{l} \left\lbrace\begin{array}{l}  \dfrac{1}{\sqrt[3]{x}}=2 \\  \dfrac{1}{\sqrt[3]{y}}=1\end{array}\right. \\ \left\lbrace\begin{array}{l}  \dfrac{1}{\sqrt[3]{x}}=1 \\  \dfrac{1}{\sqrt[3]{y}}=2\end{array}\right.\end{array}\right. \Leftrightarrow \left[\begin{array}{l} \left\lbrace\begin{array}{l} x= \dfrac{1}{8} \\ y=1\end{array}\right. \\ \left\lbrace\begin{array}{l} x=1 \\ y= \dfrac{1}{8}\end{array}\right.\end{array}\right.$$
$$S=\left\lbrace\left( \dfrac{1}{8},1\right);\left(1, \dfrac{1}{8}\right)\right\rbrace$$