Giải hệ phương trình:

a) \(\left\{ \begin{array}{l}{x^3} + {y^3} = 1\\{x^2}y + 2{\rm{x}}{y^2} + {y^3} = 2\end{array} \right.\).

b) \(\left\{ \begin{array}{l}{y^2} = \left( {x + 8} \right)\left( {{x^2} + 2} \right)\\{y^2} - 4\left( {{\rm{x + 2}}} \right)y + 16 + 16{\rm{x}} - 5{{\rm{x}}^2} = 0\end{array} \right.\).

Lời giải

a) Ta có \(\left\{ \begin{array}{l}{x^3} + {y^3} = 1\\{x^2}y + 2{\rm{x}}{y^2} + {y^3} = 2\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}2{x^3} + 2{y^3} = 2\\{x^2}y + 2{\rm{x}}{y^2} + {y^3} = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\2{{\rm{x}}^3} + {y^3} - {x^2}y - 2{\rm{x}}{y^2} = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\{{\rm{x}}^2}\left( {2{\rm{x}} - y} \right) + {y^2}\left( {y - 2{\rm{x}}} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\\left( {2{\rm{x}} - y} \right)\left( {{x^2} - {y^2}} \right) = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\\left( {2{\rm{x}} - y} \right)\left( {x - y} \right)\left( {x + y} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\\left[ \begin{array}{l}2{\rm{x}} - y = 0\\x - y = 0\\x + y = 0\end{array} \right.\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 1\\\left[ \begin{array}{l}2{\rm{x = y}}\\x = y\\x = - y\end{array} \right.\end{array} \right.\)

• \[\left\{ \begin{array}{l}{x^3} + {y^3} = 1\\y = 2{\rm{x}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + 8{{\rm{x}}^3} = 1\\y = 2{\rm{x}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \sqrt[3]{{\frac{1}{9}}}\\y = 2{\rm{x}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \sqrt[3]{{\frac{1}{9}}}\\y = 2\sqrt[3]{{\frac{1}{9}}}\end{array} \right.\]

Þ Hệ phương trình có nghiệm \(\left( {x;y} \right) = \left( {\sqrt[3]{{\frac{1}{9}}};2\sqrt[3]{{\frac{1}{9}}}} \right)\).

• \(\left\{ \begin{array}{l}{x^3} + {y^3} = 1\\x = y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{{\rm{x}}^3} = 1\\x = y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{x}} = \sqrt[3]{{\frac{1}{2}}}\\x = y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{x}} = \sqrt[3]{{\frac{1}{2}}}\\y = \sqrt[3]{{\frac{1}{2}}}\end{array} \right.\)

Þ Hệ phương trình có nghiệm \(\left( {x;y} \right) = \left( {\sqrt[3]{{\frac{1}{2}}};\sqrt[3]{{\frac{1}{2}}}} \right)\).

• \[\left\{ \begin{array}{l}{x^3} + {y^3} = 1\\x = - y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - {y^3} + {y^3} = 1\\x = - y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}0{y^3} = 1\left( {v\^o {\rm{ }}l\'i } \right)\\x = - y\end{array} \right.\]

Vậy hệ phương trình đã cho có tập nghiệm là \(S = \left\{ {\left( {\sqrt[3]{{\frac{1}{9}}};2\sqrt[3]{{\frac{1}{9}}}} \right);\left( {\sqrt[3]{{\frac{1}{2}}};\sqrt[3]{{\frac{1}{2}}}} \right)} \right\}\).

b) \(\left\{ \begin{array}{l}{y^2} = \left( {x + 8} \right)\left( {{x^2} + 2} \right)\\{y^2} - 4\left( {{\rm{x + 2}}} \right)y + 16 + 16{\rm{x}} - 5{{\rm{x}}^2} = 0\end{array} \right.\)

Ta có

⇔ \(\left[ \begin{array}{l}x + y - 4 = 0\\y - 5{\rm{x}} - 4 = 0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}y = 4 - x\\y = 5{\rm{x + }}4\end{array} \right.\)

+) Nếu y = 5x + 4

y² = (x + 8)(x² + 2)

⇔ (5x + 4)² = (x + 8)(x² + 2)

⇔ 25x² – 40x + 16 = x³ + 2x + 8x² + 16

⇔ x³ – 17x² + 42x = 0

⇔ x(x² – 17x + 42) = 0

⇔ x(x – 3)(x – 14) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x = 0 \Rightarrow y = 4\\x = 3 \Rightarrow y = 19\\x = 14 \Rightarrow y = 74\end{array} \right.\)

+) Nếu y = 4 – x

y² = (x + 8)(x² + 2)

⇔ (4 – x)² = (x + 8)(x² + 2)

⇔ x² – 8x + 16 = x³ + 2x + 8x² + 16

⇔ x³ + 7x² + 10x = 0

⇔ x(x² + 7x + 10) = 0

⇔ x(x + 2)(x + 5) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x = 0 \Rightarrow y = 4\\x = - 2 \Rightarrow y = 6\\x = - 5 \Rightarrow y = 9\end{array} \right.\)

Vậy hệ phương trình đã cho có tập nghiệm là S = {(0; 4); (3; 19); (14; 74); (0; 4); (–2; 6); (–5; 9)}.