giải hpt : 1) x=2y/1-y^2 và y=2x/1-x^2
2) x^3=5x+y và y^3=5y+x
giải hpt : 1) x=2y/1-y^2 và y=2x/1-x^2 2) x^3=5x+y và y^3=5y+x
By Samantha
By Samantha
giải hpt : 1) x=2y/1-y^2 và y=2x/1-x^2
2) x^3=5x+y và y^3=5y+x
Đáp án:
1) $S = \left\{ {\left( {0;0} \right),\left( {\sqrt {15} ;\dfrac{{\sqrt {15} }}{5}} \right),\left( { – \sqrt {15} ;\dfrac{{ – \sqrt {15} }}{5}} \right)} \right\}$
2) $S = \left\{ {\left( {0;0} \right),\left( {\sqrt 6 ;\sqrt 6 } \right),\left( { – \sqrt 6 ; – \sqrt 6 } \right),\left( {2; – 2} \right),\left( { – 2;2} \right),\left( {\dfrac{{\sqrt 7 – \sqrt 3 }}{2};\dfrac{{ – \sqrt 7 – \sqrt 3 }}{2}} \right),\left( {\dfrac{{ – \sqrt 7 – \sqrt 3 }}{2};\dfrac{{\sqrt 7 – \sqrt 3 }}{2}} \right),\left( {\dfrac{{\sqrt 7 + \sqrt 3 }}{2};\dfrac{{ – \sqrt 7 + \sqrt 3 }}{2}} \right),\left( {\dfrac{{ – \sqrt 7 + \sqrt 3 }}{2};\dfrac{{\sqrt 7 + \sqrt 3 }}{2}} \right)} \right\}$
Giải thích các bước giải:
1) ĐKXĐ: $x,y\ne \pm 1$
Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
x = \dfrac{{2y}}{{1 – {y^2}}}\\
y = \dfrac{{2x}}{{1 – {x^2}}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 – {y^2}} \right) = 2y\\
y\left( {1 – {x^2}} \right) = 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 – {y^2}} \right) = 2y\\
x\left( {1 – {y^2}} \right) – y\left( {1 – {x^2}} \right) = 2\left( {y – x} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 – {y^2}} \right) = 2y\\
x – y – x{y^2} + {x^2}y + 2\left( {x – y} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 – {y^2}} \right) = 2y\\
\left( {x – y} \right)\left( {3 – xy} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 – {y^2}} \right) = 2y\left( 1 \right)\\
\left[ \begin{array}{l}
x = y\\
xy = 3
\end{array} \right.
\end{array} \right.\\
+ )TH1:x = y\\
\left( 1 \right)tt:y\left( {1 – {y^2}} \right) = 2y\\
\Leftrightarrow y\left( {{y^2} + 1} \right) = 1\\
\Leftrightarrow y = 0 \Rightarrow x = y = 0\left( {tm} \right)\\
+ )TH2:xy = 3 \Rightarrow x,y \ne 0\\
\left( 1 \right) \Leftrightarrow x – x{y^2} = 2y\\
\Leftrightarrow x – xy.y = 2y\\
\Leftrightarrow x – 3y = 2y\\
\Leftrightarrow x = 5y\\
\Leftrightarrow \dfrac{3}{y} = 5y\\
\Leftrightarrow {y^2} = \dfrac{3}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
y = \dfrac{{\sqrt {15} }}{5}\\
y = \dfrac{{ – \sqrt {15} }}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt {15} ;y = \dfrac{{\sqrt {15} }}{5}\\
x = – \sqrt {15} ;y = \dfrac{{ – \sqrt {15} }}{5}
\end{array} \right.\left( {tm} \right)\\
\end{array}$
Vậy hệ có tập nghiệm: $S = \left\{ {\left( {0;0} \right),\left( {\sqrt {15} ;\dfrac{{\sqrt {15} }}{5}} \right),\left( { – \sqrt {15} ;\dfrac{{ – \sqrt {15} }}{5}} \right)} \right\}$
2) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^3} = 5x + y\\
{y^3} = 5y + x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
{x^3} – {y^3} = 5x + y – \left( {5y + x} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
{x^3} – {y^3} – 4\left( {x – y} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
\left( {x – y} \right)\left( {{x^2} + xy + {y^2} – 4} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\left( 1 \right)\\
\left[ \begin{array}{l}
x – y = 0\\
{x^2} + xy + {y^2} – 4 = 0
\end{array} \right.
\end{array} \right.\\
+ )TH1:x – y = 0 \Leftrightarrow x = y\\
\left( 1 \right)tt:{x^3} = 6x\\
\Leftrightarrow {x^3} – 6x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt 6 \\
x = – \sqrt 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = \sqrt 6 \\
x = y = – \sqrt 6
\end{array} \right.\\
+ )TH2:{x^2} + xy + {y^2} – 4 = 0\\
\left\{ \begin{array}{l}
{x^3} = 5x + y\\
{x^2} + xy + {y^2} – 4 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
{x^2} + xy + {y^2} = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
4{x^3} = \left( {5x + y} \right)\left( {{x^2} + xy + {y^2}} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
{x^3} + {y^3} + 6{x^2}y + 6x{y^2} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
\left( {x + y} \right)\left( {{x^2} + 5xy + {y^2}} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^3} = 5x + y\\
\left[ \begin{array}{l}
y = – x\\
{x^2} + 5xy + {y^2} = 0
\end{array} \right.
\end{array} \right.\\
*)y = – x\\
\left( 1 \right)tt:{x^3} = 5x – x\\
\Leftrightarrow {x^3} – 4x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = – 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = y = 0\\
x = 2;y = – 2\\
x = – 2;y = 2
\end{array} \right.\\
*){x^2} + 5xy + {y^2} = 0\\
\left\{ \begin{array}{l}
{x^2} + 5xy + {y^2} = 0\\
{x^2} + xy + {y^2} = 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
xy = – 1\\
{x^2} + {y^2} = 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
xy = – 1\\
{\left( {x + y} \right)^2} – 2xy = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
xy = – 1\\
{\left( {x + y} \right)^2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
xy = – 1\\
\left[ \begin{array}{l}
x + y = \sqrt 3 \\
x + y = – \sqrt 3
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + y = \sqrt 3 \\
xy = – 1
\end{array} \right.\left( I \right)\\
\left\{ \begin{array}{l}
x + y = – \sqrt 3 \\
xy = – 1
\end{array} \right.\left( {II} \right)
\end{array} \right.\\
\left( I \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt 7 + \sqrt 3 }}{2};y = \dfrac{{ – \sqrt 7 + \sqrt 3 }}{2}\\
x = \dfrac{{ – \sqrt 7 + \sqrt 3 }}{2};y = \dfrac{{\sqrt 7 + \sqrt 3 }}{2}
\end{array} \right.\\
\left( {II} \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt 7 – \sqrt 3 }}{2};y = \dfrac{{ – \sqrt 7 – \sqrt 3 }}{2}\\
x = \dfrac{{ – \sqrt 7 – \sqrt 3 }}{2};y = \dfrac{{\sqrt 7 – \sqrt 3 }}{2}
\end{array} \right.
\end{array}$
Vậy hệ có tập nghiệm là:
$S = \left\{ {\left( {0;0} \right),\left( {\sqrt 6 ;\sqrt 6 } \right),\left( { – \sqrt 6 ; – \sqrt 6 } \right),\left( {2; – 2} \right),\left( { – 2;2} \right),\left( {\dfrac{{\sqrt 7 – \sqrt 3 }}{2};\dfrac{{ – \sqrt 7 – \sqrt 3 }}{2}} \right),\left( {\dfrac{{ – \sqrt 7 – \sqrt 3 }}{2};\dfrac{{\sqrt 7 – \sqrt 3 }}{2}} \right),\left( {\dfrac{{\sqrt 7 + \sqrt 3 }}{2};\dfrac{{ – \sqrt 7 + \sqrt 3 }}{2}} \right),\left( {\dfrac{{ – \sqrt 7 + \sqrt 3 }}{2};\dfrac{{\sqrt 7 + \sqrt 3 }}{2}} \right)} \right\}$