Đề thi online - Hệ phương trình đối xứng - Có lời giải chi tiết

Câu 1 : \(\left\{ \begin{array}{l}{x^2} + x = {y^2} + y\\{x^2} + {y^2} = 5\end{array} \right.(x,y \in R)\)

A \(\left( {1;\,1} \right),\,\,\left( { - 1; - 1} \right),\left( {\frac{{\sqrt {10} }}{2};\frac{{ - \sqrt {10} - 2}}{2}} \right),\left( { - \frac{{\sqrt {10} }}{2};\frac{{\sqrt {10} - 2}}{2}} \right)\)

B \(\left( {1; - 2} \right),\,\,\left( { - 2;1} \right),\left( {\frac{{\sqrt {10} }}{2};\frac{{\sqrt {10} }}{2}} \right),\left( { - \frac{{\sqrt {10} }}{2}; - \frac{{\sqrt {10} }}{2}} \right)\)

C \(\left( { - 1;0} \right),\,\,\left( {2; - 1} \right),\left( {\frac{{\sqrt {10} }}{2}; - \frac{{\sqrt {10} }}{2}} \right),\left( { - \frac{{\sqrt {10} }}{2};\frac{{\sqrt {10} }}{2}} \right)\)

D \(\left( { - 1;0} \right),\,\,\left( {2; - 1} \right),\left( { - \frac{{\sqrt {10} }}{2};\frac{{\sqrt {10} - 2}}{2}} \right),\left( {\frac{{\sqrt {10} }}{2};\frac{{ - \sqrt {10} - 2}}{2}} \right)\)

Câu 2 : \(\left\{ \begin{array}{l}\frac{1}{x} + \frac{1}{y} = - \frac{1}{2}\\{x^2} + {y^2} = 5\end{array} \right.{\rm{ }}\left( * \right)\)

A \(S = \left( {x;y} \right) = \left\{ {\left( {1;0} \right),\left( { - 2;3} \right)} \right\}\)

B \(S = \left( {x;y} \right) = \left\{ {\left( { - 1;2} \right),\left( {2; - 1} \right)} \right\}\)

C \(S = \left( {x;y} \right) = \left\{ {\left( { - 1; - 2} \right),\left( { - 2; - 1} \right)} \right\}\)

D \(S = \left( {x;y} \right) = \left\{ {\left( {1; - 2} \right),\left( { - 2;1} \right)} \right\}\)

Câu 3 : Giải hệ phương trình \(\left\{ \begin{array}{l}2{x^2} - 3x = {y^2} - 2\,\,\,\,\left( * \right)\\2{y^2} - 3y = {x^2} - 2\,\,\,\left( {**} \right)\end{array} \right.\)

A \(\left( {x;\,y} \right) = \left\{ {\left( {1;\,1} \right);\,\left( {2;\,2} \right)} \right\}\)

B \(\left( {x;\,y} \right) = \left\{ {\left( { - 1; - \,1} \right);\,\left( { - 2;\, - 2} \right)} \right\}\)

C \(\left( {x;\,y} \right) = \left\{ {\left( { - 1;\,1} \right);\,\left( {2;\, - 2} \right)} \right\}\)

D \(\left( {x;\,y} \right) = \left\{ {\left( {1;\, - 1} \right);\,\left( { - 2;\,2} \right)} \right\}\)

Câu 4 : Giải hệ phương trình \(\left\{ \begin{array}{l}{x^3} = 3x + 8y\,\,\,\,\left( * \right)\\{y^3} = 3y + 8x{\rm{ }}\left( {**} \right)\end{array} \right.\)

A \(\left( {x;\,y} \right) = \left\{ {\left( {0;0} \right);\,\,\left( {\sqrt {11} ; - \sqrt {11} } \right);\,\,\left( { - \sqrt {11} ;\sqrt {11} } \right)} \right\}\)

B \(\left( {x;\,y} \right) = \left\{ {\left( {0;0} \right);\,\,\left( {\sqrt {13} ;\sqrt {13} } \right);\,\,\left( { - \sqrt {13} ; - \sqrt {13} } \right)} \right\}\)

C \(\left( {x;\,y} \right) = \left\{ {\left( {0;0} \right);\,\,\left( {\sqrt {13} ; - \sqrt {13} } \right);\,\,\left( { - \sqrt {13} ;\sqrt {13} } \right)} \right\}\)

D \(\left( {x;\,y} \right) = \left\{ {\left( {0;0} \right);\,\,\left( {\sqrt {11} ;\sqrt {11} } \right);\,\,\left( { - \sqrt {11} ; - \sqrt {11} } \right)} \right\}\)

Câu 5 : Giải hệ phương trình: \(\left\{ \begin{array}{l}\sqrt {2x + 3} + \sqrt {4 - y} = 4{\rm{ }}\left( 1 \right)\\\sqrt {2y + 3} + \sqrt {4 - x} = 4{\rm{ }}\left( 2 \right)\end{array} \right.\)

A \(S = \left( {x;y} \right) = \left\{ {\left( { - 3; - 3} \right),\left( {\frac{{11}}{9};\frac{{11}}{9}} \right)} \right\}\)

B \(S = \left( {x;y} \right) = \left\{ {\left( {3;3} \right),\left( { - \frac{{11}}{9}; - \frac{{11}}{9}} \right)} \right\}\)

C \(S = \left( {x;y} \right) = \left\{ {\left( {3;3} \right),\left( {\frac{{11}}{9};\frac{{11}}{9}} \right)} \right\}\)

D \(S = \left( {x;y} \right) = \left\{ {\left( { - 3; - 3} \right),\left( { - \frac{{11}}{9}; - \frac{{11}}{9}} \right)} \right\}\)

Câu 6 : Tìm \(m\) để hệ phương trình sau có nghiệm: \(\left\{ \begin{array}{l}x + \frac{1}{x} + y + \frac{1}{y} = 5\\{x^3} + \frac{1}{{{x^3}}} + {y^3} + \frac{1}{{{y^3}}} = 15m - 10\end{array} \right.\)

A \(m \ge 22\)

B \(m \le 22\)

C \(m \ge 11\)

D \(m \le 11\)

Câu 7 : \(\left\{ \begin{array}{l}{x^2} + y + \frac{1}{4} = 0\\x + {y^2} + \frac{1}{4} = 0\end{array} \right.\)

A \(\left( {x;y} \right) = \left( {\frac{1}{2};\frac{1}{2}} \right)\)

B \(\left( {x;y} \right) = \left( {\frac{{ - 1}}{2};\frac{{ - 1}}{2}} \right)\)

C \(\left( {x;y} \right) = \left( {\frac{1}{2};\frac{{ - 1}}{2}} \right)\)

D \(\left( {x;y} \right) = \left( {\frac{{ - 1}}{2};\frac{1}{2}} \right)\)

Câu 8 : \(\left\{ \begin{array}{l}2{x^2} = y + \frac{1}{y}\\2{y^2} = x + \frac{1}{x}\end{array} \right.\)

A \(\left( {x;y} \right) = \left( {1;1} \right)\)

B \(\left( {x;y} \right) = \left( { - 1; - 1} \right)\)

C \(\left( {x;y} \right) = \left( { - 1;1} \right)\)

D \(\left( {x;y} \right) = \left( {1; - 1} \right)\)

Câu 9 : Giải hệ phương trình \(\left\{ \begin{array}{l}{x^2} + {y^2} = 65\\\left( {x - 1} \right)\left( {y - 1} \right) = 18\end{array} \right.\)

A \(\left( {1;8} \right);\left( {8;1} \right);\left( {4;7} \right);\left( {7;4} \right)\)

B \(\left( { - 1; - 8} \right);\left( { - 8; - 1} \right);\left( { - 4; - 7} \right);\left( { - 7; - 4} \right)\)

C \(\left( { - 1; - 8} \right);\left( { - 8; - 1} \right);\left( {4;7} \right);\left( {7;4} \right)\)

D \(\left( {1;8} \right);\left( {8;1} \right);\left( { - 4; - 7} \right);\left( { - 7; - 4} \right)\)

Câu 10 : Giải hệ phương trình : \(\left\{ \begin{array}{l}\left( {x + y} \right)\left( {1 + \frac{1}{{xy}}} \right) = 5\\\left( {{x^2} + {y^2}} \right)\left( {1 + \frac{1}{{{x^2}{y^2}}}} \right) = 9\end{array} \right.\)

A \(\left( {x;\,y} \right) = \left\{ {\left( {1;\frac{{3 - \sqrt 5 }}{2}} \right),\left( {1;\frac{{3 + \sqrt 5 }}{2}} \right),\left( {\frac{{3 - \sqrt 5 }}{2}; - 1} \right),\left( {\frac{{3 + \sqrt 5 }}{2}; - 1} \right)} \right\}\)

B \(\left( {x;\,y} \right) = \left\{ {\left( {1; - \frac{{3 - \sqrt 5 }}{2}} \right),\left( {1; - \frac{{3 + \sqrt 5 }}{2}} \right),\left( {\frac{{3 - \sqrt 5 }}{2};1} \right),\left( {\frac{{3 + \sqrt 5 }}{2};1} \right)} \right\}\)

C \(\left( {x;\,y} \right) = \left\{ {\left( { - 1; - \frac{{3 - \sqrt 5 }}{2}} \right),\left( {1;\frac{{3 + \sqrt 5 }}{2}} \right),\left( {\frac{{3 - \sqrt 5 }}{2};1} \right),\left( { - \frac{{3 + \sqrt 5 }}{2}; - 1} \right)} \right\}\)

D \(\left( {x;\,y} \right) = \left\{ {\left( {1;\frac{{3 - \sqrt 5 }}{2}} \right),\left( {1;\frac{{3 + \sqrt 5 }}{2}} \right),\left( {\frac{{3 - \sqrt 5 }}{2};1} \right),\left( {\frac{{3 + \sqrt 5 }}{2};1} \right)} \right\}\)

Câu 11 : Giả sử \(\left( {x;y} \right)\) là nghiệm của hệ phương trình \(\left\{ \begin{array}{l}x + y = 2a - 1\\{x^2} + {y^2} = {a^2} + 2a - 3\end{array} \right.\). Xác định \(a\) để tích \(xy\) đạt giá trị nhỏ nhất. Tìm giá trị đó .

A \(a = \frac{{4 - \sqrt 2 }}{2}\)

B \(a = \frac{{4 + \sqrt 2 }}{2}\)

C \(a = \frac{{ - 4 + \sqrt 2 }}{2}\)

D \(a = \frac{{ - 4 - \sqrt 2 }}{2}\)

Câu 12 : Giải hệ phương trình \(\left\{ \begin{array}{l}{x^2} = 2x - y\\{y^2} = 2y - z\\{z^2} = 2z - t\\{t^2} = 2t - x\end{array} \right.\)

A \(\left( {x;\,y;\,z;\,t} \right) = \left\{ {\left( {0;0;0;0} \right);\left( { - 1; - 1; - 1; - 1} \right)} \right\}\)

B \(\left( {x;\,y;\,z;\,t} \right) = \left\{ {\left( {0;0;0;0} \right);\left( {2;2;2;2} \right)} \right\}\)

C \(\left( {x;\,y;\,z;\,t} \right) = \left\{ {\left( {1;1;1;1} \right);\left( {2;2;2;2} \right)} \right\}\)

D \(\left( {x;\,y;\,z;\,t} \right) = \left\{ {\left( {0;0;0;0} \right);\left( {1;1;1;1} \right)} \right\}\)

Đề thi online - Hệ phương trình đối xứng - Có lời...