$$\left\lbrace\begin{array}{l} (x-y)^2(3x^2+2xy+3y^2-20)+1=0\\2x^2-5x-2xy+5y=0 \end{array}\right.$$

Đề bài:

\begin{equation} \label{eq:74.I} \left\lbrace\begin{array}{l} (x-y)^2(3x^2+2xy+3y^2-20)+1=0\\2x^2-5x-2xy+5y=0 \end{array}\right. \end{equation}

Lời giải:

Xét $x = y, \eqref{eq:74.I} \Rightarrow \left\lbrace\begin{array}{l} 1=0 \\ 0=0 \end{array}\right.$ (vô lý)
Xét $x\neq y$
\begin{align} \eqref{eq:74.I} &\Leftrightarrow \left\lbrace\begin{array}{l} (x-y)^2(3x^2+2xy+3y^2-20)+1=0\\(2x-5)(x-y)=0 \end{array}\right.\nonumber \\ &\Leftrightarrow \left\lbrace\begin{array}{l} (3x^2+2xy+3y^2-20)+ \dfrac{1}{(x-y)^2}=0\\2x=5 \end{array}\right.\nonumber \\ &\Leftrightarrow \left\lbrace\begin{array}{l} x^2-2xy+y^2+2x^2+4xy+2y^2+ \dfrac{1}{(x-y)^2}=20\\x+y+x-y=5 \end{array}\right.\nonumber \\ \label{eq:74.1} &\Leftrightarrow \left\lbrace\begin{array}{l} (x-y)^2+2(x+y)^2+ \dfrac{1}{(x-y)^2}=20\\x+y+x-y=5 \end{array}\right.\end{align}
Đặt $x+y=a$; $x-y=b$, ta có:
\begin{align}\eqref{eq:74.1} & \Rightarrow \left\lbrace\begin{array}{l} b^2+2a^2+ \dfrac{1}{b^2}=20 \\ a+b=5 \end{array}\right. \nonumber \\& \Leftrightarrow \left\lbrace\begin{array}{l} b^2+2(5-b)^2+ \dfrac{1}{b^2}=20 \\ a=5-b \end{array}\right. \nonumber \\& \Leftrightarrow \left\lbrace\begin{array}{l} b^2+50-20b+2b^2+ \dfrac{1}{b^2}=20 \\ a=5-b \end{array}\right. \nonumber \\& \Leftrightarrow \left\lbrace\begin{array}{l} 3b^2-20b+30+ \dfrac{1}{b^2}=0 \\ a=5-b \end{array}\right. \nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} 3b^4-20b^3+30b^2+1=0 \\ a=5-b \end{array} \right. \nonumber \end{align}
Xét phương trình \begin{equation} \label{eq:74.2} 3b^4-20b^3+30b^2+1=0 \end{equation}
\begin{align} \eqref{eq:74.2} & \Leftrightarrow 9b^4-60b^3+90b^2+3=0 \nonumber \\ & \Leftrightarrow \left(3b^2\right)^2-2.3b^2.10b+(10b)^2-10b^2+3=0 \nonumber \\\label{eq:74.3} & \Leftrightarrow \left(3b^2-10b\right)^2=10b^2-3 \end{align}
Đưa thêm vào một ẩn mới là $n$, cộng hai vế của \eqref{eq:74.3} với $\left( 3b^2-10b\right)n+ \dfrac{n^2}{4}$, ta có:
\begin{equation} \label{eq:74.4} \eqref{eq:74.3} \Leftrightarrow \left(3b^2-10b+ \dfrac{n}{2}\right)^2=(3n+10)b^2-10nb+\left( \dfrac{n^2}{4}-3\right) \end{equation}
Chọn $n$ để vế phải của \eqref{eq:74.4} là một bình phương, hay là biệt thức của đa thức vế phải với biến $b$ bằng 0, cụ thể là:
\begin{align} \label{eq:74.8} & \phantom{\Leftrightarrow} 100n^2-(3n+10)\left( n^2-12\right)=0 \\ & \Leftrightarrow -3n^3+90n^2+36n+120=0 \nonumber \\ \label{eq:74.5} & \Leftrightarrow n^3-30n^2-12n-40=0 \end{align}
Đặt $n=m+10$
\begin{align} \eqref{eq:74.5} & \Rightarrow (m+10)^3-30(m+10)^2-12(m+10)-40=0 \nonumber \\ & \Leftrightarrow m^3+30m^2+300m+1000 -30m^2-600m-3000-12m-120-40=0 \nonumber \\ \label{eq:74.6} & \Leftrightarrow m^3-312m-2160=0 \end{align}
Đặt $m=u+v$ với $(u\geqslant v)$, ta có:
\begin{align} \eqref{eq:74.6} & \Rightarrow (u+v)^3-312(u+v)-2160=0 \nonumber \\ & \Leftrightarrow u^3+v^3+3uv(u+v)-312(u+v)-2160=0 \nonumber \\ \label{eq:74.7} & \Leftrightarrow u^3+v^3-2160+(3uv-312)(u+v)=0 \end{align}
Chọn $u$, $v$ sao cho $3uv-312=0$, khi đó ta có hệ phương trình sau:
\begin{align} & \phantom{\Leftrightarrow} \left\lbrace \begin{array}{l} u^3+v^3=2160 \\ 3uv-312=0 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} u^3+v^3=2160 \\ uv=104 \end{array} \right. \nonumber \\ & \Leftrightarrow \left\lbrace \begin{array}{l} u^3+v^3=2160 \\ u^3v^3=1124864 \end{array} \right. \nonumber \end{align}
Áp dụng định lý $Viete$ đảo ta có $u^3,v^3$ là nghiệm của phương trình
\begin{align} & m^2-2160m+1124864=0 \quad \left(\Delta'=\left( \dfrac{-2160}{2}\right)^2-1124864=41536\right) \nonumber \\ & \Rightarrow \left\lbrace \begin{array}{l} u^3=1080+8\sqrt{649} \\ v^3=1080-8\sqrt{649} \end{array} \right. \Leftrightarrow \left\lbrace \begin{array}{l} u=\sqrt[3]{1080+8\sqrt{649}} \\ v=\sqrt[3]{1080-8\sqrt{649}} \end{array} \right. \nonumber \\ & \Rightarrow m=u+v=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}} \nonumber \\ & \Rightarrow n=m+10=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}}+10 \nonumber \end{align}
Giá trị của $n$ mà ta mới tìm được ở trên chính là giá trị $n$ thỏa mãn \eqref{eq:74.8}, tức là
$$100n^2-(3n+10)\left( n^2-12\right)=0 \Rightarrow 2\sqrt{3n+10}\sqrt{ \dfrac{n^2}{4}-3} = 10n$$
Khi đó, ta có:
\begin{align} \eqref{eq:74.4} & \Leftrightarrow \left(3b^2-10b+ \dfrac{n}{2}\right)^2=(3n+10)b^2-2\sqrt{3n+10}\sqrt{ \dfrac{n^2}{4}-3}.b+\left( \dfrac{n^2}{4}-3\right) \nonumber \\ & \Leftrightarrow \left(3b^2-10b+ \dfrac{n}{2}\right)^2=\left( \sqrt{3n+10}b-\sqrt{ \dfrac{n^2}{4}-3}\right)^2 \nonumber \\ & \Leftrightarrow \left(3b^2-10b+ \dfrac{n}{2}\right)^2-\left( \sqrt{3n+10}b-\sqrt{ \dfrac{n^2}{4}-3}\right)^2=0 \nonumber \\& \Leftrightarrow \left[3b^2-\left(10+\sqrt{3n+10}\right)b+\left( \dfrac{n}{2}+\sqrt{ \dfrac{n^2}{4}-3}\right)\right] \times \nonumber \\ & \hspace{5cm} \times \left[3b^2-\left(10-\sqrt{3n+10}\right)b+\left( \dfrac{n}{2}-\sqrt{ \dfrac{n^2}{4}-3}\right)\right]=0 \nonumber \end{align}
Hệ phương trình trên tương đương với hệ hai phương trình sau:
\begin{align} \label{eq:74.9} 3b^2-\left(10+\sqrt{3n+10}\right)b+\left( \dfrac{n}{2}+\sqrt{ \dfrac{n^2}{4}-3}\right)=0 \\ \label{eq:74.10} 3b^2-\left(10-\sqrt{3n+10}\right)b+\left( \dfrac{n}{2}-\sqrt{ \dfrac{n^2}{4}-3}\right)=0 \end{align}
Ta giải \eqref{eq:74.9}.
\begin{align} \Delta & =\left(10+\sqrt{3n+10}\right)^2-6n-12\sqrt{ \dfrac{n^2}{4}-3} \nonumber \\ & =\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12} \nonumber \end{align}
Với giá trị $n=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}}+10$ thì $\Delta>0$
Khi đó ta có:
\begin{align} & \left[ \begin{array}{l} b= \dfrac{10+\sqrt{3n+10}+\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \\ b= \dfrac{10+\sqrt{3n+10}-\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \end{array} \right. \nonumber \\ \Rightarrow & \left[ \begin{array}{l} a=5-b= \dfrac{20-\sqrt{3n+10}-\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \\ a=5-b= \dfrac{20-\sqrt{3n+10}+\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \end{array} \right. \nonumber \\ \Rightarrow & \left[ \begin{array}{l} \left\lbrace \begin{array}{l} x+y= \dfrac{20-\sqrt{3n+10}-\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \\ x-y= \dfrac{10+\sqrt{3n+10}+\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \end{array} \right. \\ \left\lbrace \begin{array}{l} x+y= \dfrac{20-\sqrt{3n+10}+\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \\ x-y= \dfrac{10+\sqrt{3n+10}-\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6}\end{array} \right. \end{array} \right. \nonumber \\ \Rightarrow & \left[ \begin{array}{l} \left\lbrace \begin{array}{l} x= \dfrac{5}{2} \\ y= \dfrac{5-\sqrt{3n+10}-\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \end{array} \right. \\ \left\lbrace \begin{array}{l} x= \dfrac{5}{2} \\ y= \dfrac{5-\sqrt{3n+10}+\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \end{array} \right. \end{array} \right. \nonumber \end{align}
(với $n=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}}+10$)
Ta giải \eqref{eq:74.10}.
\begin{align} \Delta & =\left(10-\sqrt{3n+10}\right)^2-6n+12\sqrt{ \dfrac{n^2}{4}-3} \nonumber \\ & =\left(10-\sqrt{3n+10}\right)^2-6n+6\sqrt{n^2-12} \nonumber \end{align}
Với giá trị $n=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}}+10$ thì $\Delta<0$. Khi đó \eqref{eq:74.10} sẽ vô nghiệm.
$$S=\left\lbrace \left( \dfrac{5}{2}; \dfrac{5-\sqrt{3n+10}\pm\sqrt{\left(10+\sqrt{3n+10}\right)^2-6n-6\sqrt{n^2-12}}}{6} \right) \right\rbrace $$
(với $n=\sqrt[3]{1080+8\sqrt{649}}+\sqrt[3]{1080-8\sqrt{649}}+10$)