giải hệ pt (x-1)(y-1)(x+y-2)=6 x^2+y^2-2x-2y-3=0 19/08/2021 Bởi Serenity giải hệ pt (x-1)(y-1)(x+y-2)=6 x^2+y^2-2x-2y-3=0
Đáp án: \[\left[ \begin{array}{l}x = 2;\,\,y = 3\\x = 3;\,\,y = 2\end{array} \right.\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\left\{ \begin{array}{l}\left( {x – 1} \right)\left( {y – 1} \right)\left( {x + y – 2} \right) = 6\\{x^2} + {y^2} – 2x – 2y – 3 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {x – 1} \right)\left( {y – 1} \right)\left( {\left( {x – 1} \right) + \left( {y – 1} \right)} \right) = 6\\\left( {{x^2} – 2x + 1} \right) + \left( {{y^2} – 2y + 1} \right) – 5 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left( {x – 1} \right)\left( {y – 1} \right)\left( {\left( {x – 1} \right) + \left( {y – 1} \right)} \right) = 6\\{\left( {x – 1} \right)^2} + {\left( {y – 1} \right)^2} = 5\end{array} \right.\end{array}\) Đặt \(a = x – 1;\,\,\,b = y – 1\), khi đó, hệ phương trình trên trở thành: \(\begin{array}{l}\left\{ \begin{array}{l}ab.\left( {a + b} \right) = 6\\{a^2} + {b^2} = 5\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\{\left( {a + b} \right)^2} – 2ab = 5\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a,b \ne 0} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\{\left( {a + b} \right)^2} – 2.\dfrac{6}{{a + b}} = 5\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\{\left( {a + b} \right)^3} – 12 = 5\left( {a + b} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\{\left( {a + b} \right)^3} – 5\left( {a + b} \right) – 12 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\\left[ {{{\left( {a + b} \right)}^3} – 3.{{\left( {a + b} \right)}^2}} \right] + \left[ {3.{{\left( {a + b} \right)}^2} – 9\left( {a + b} \right)} \right] + \left[ {4.\left( {a + b} \right) – 12} \right] = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\\left[ {\left( {a + b} \right) – 3} \right]\left[ {{{\left( {a + b} \right)}^2} + 3.\left( {a + b} \right) + 4} \right] = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}ab = \dfrac{6}{{a + b}}\\a + b = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a + b = 3\\ab = 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3 – a\\a.\left( {3 – a} \right) = 2\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3 – a\\{a^2} – 3a + 2 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3 – a\\\left( {a – 1} \right)\left( {a – 2} \right) = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}a = 1;\,\,b = 2\\a = 2;\,\,b = 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 2;\,\,y = 3\\x = 3;\,\,y = 2\end{array} \right.\end{array}\) Bình luận
Đáp án:
\[\left[ \begin{array}{l}
x = 2;\,\,y = 3\\
x = 3;\,\,y = 2
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {x – 1} \right)\left( {y – 1} \right)\left( {x + y – 2} \right) = 6\\
{x^2} + {y^2} – 2x – 2y – 3 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x – 1} \right)\left( {y – 1} \right)\left( {\left( {x – 1} \right) + \left( {y – 1} \right)} \right) = 6\\
\left( {{x^2} – 2x + 1} \right) + \left( {{y^2} – 2y + 1} \right) – 5 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x – 1} \right)\left( {y – 1} \right)\left( {\left( {x – 1} \right) + \left( {y – 1} \right)} \right) = 6\\
{\left( {x – 1} \right)^2} + {\left( {y – 1} \right)^2} = 5
\end{array} \right.
\end{array}\)
Đặt \(a = x – 1;\,\,\,b = y – 1\), khi đó, hệ phương trình trên trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
ab.\left( {a + b} \right) = 6\\
{a^2} + {b^2} = 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^2} – 2ab = 5
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {a,b \ne 0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^2} – 2.\dfrac{6}{{a + b}} = 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^3} – 12 = 5\left( {a + b} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
{\left( {a + b} \right)^3} – 5\left( {a + b} \right) – 12 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
\left[ {{{\left( {a + b} \right)}^3} – 3.{{\left( {a + b} \right)}^2}} \right] + \left[ {3.{{\left( {a + b} \right)}^2} – 9\left( {a + b} \right)} \right] + \left[ {4.\left( {a + b} \right) – 12} \right] = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
\left[ {\left( {a + b} \right) – 3} \right]\left[ {{{\left( {a + b} \right)}^2} + 3.\left( {a + b} \right) + 4} \right] = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
ab = \dfrac{6}{{a + b}}\\
a + b = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a + b = 3\\
ab = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 3 – a\\
a.\left( {3 – a} \right) = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 3 – a\\
{a^2} – 3a + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = 3 – a\\
\left( {a – 1} \right)\left( {a – 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a = 1;\,\,b = 2\\
a = 2;\,\,b = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2;\,\,y = 3\\
x = 3;\,\,y = 2
\end{array} \right.
\end{array}\)