8. Bài tập ôn tập chương 3

\(\left\{ \begin{array}{l}x + ay = 1\\ – ax + y = a\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1 – ay\\ – a(1 – ay) + y = a\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1 – ay\\y({a^2} + 1) = 2a\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1 – ay\\y = \dfrac{{2a}}{{{a^2} + 1}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{1 – {a^2}}}{{{a^2} + 1}}\\y = \dfrac{{2a}}{{{a^2} + 1}}\end{array} \right.\)

Để hệ phương trình đã cho có nghiệm thỏa mãn: \(x < 1;y < 1\)  \( \Rightarrow \left\{ \begin{array}{l}\dfrac{{1 – {a^2}}}{{{a^2} + 1}} < 1\\\dfrac{{2a}}{{{a^2} + 1}} < 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 – {a^2} < {a^2} + 1\\2a < {a^2} + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{a^2} > 0\\(a – 1){}^2 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ne 0\\a \ne 1\end{array} \right.\)

+) Xét \(y = 0\) hệ phương trình đã cho trở thành: \(\left\{ \begin{array}{l}{x^2} + 1 = 0\\({x^2} + 1)(x – 2) = 0\end{array} \right.\)  (vô lí).

+) Xét \(y \ne 0\) chia các vế của từng phương trình cho y, ta được: \(\left\{ \begin{array}{l}\dfrac{{{x^2} + 1}}{y} + y + x = 4\\\dfrac{{{x^2} + 1}}{y}(y + x – 2) = 1\end{array} \right.\)

Đặt: \(\left\{ \begin{array}{l}\dfrac{{{x^2} + 1}}{y} = a\\y + x – 2 = b\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a + b = 2\\ab = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 2 – a\\a(2 – a) = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 2 – a\\{a^2} – 2a + 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 2 – a\\{(a – 1)^2} = 0\end{array} \right.\)   

\(\begin{array}{l} \Leftrightarrow a = b = 1 \Rightarrow \left\{ \begin{array}{l}\dfrac{{{x^2} + 1}}{y} = 1\\y + x – 2 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = {x^2} + 1\\x + y = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = {x^2} + 1\\x + {x^2} + 1 = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = {x^2} + 1\\{x^2} + x – 2 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}y = {x^2} + 1\\(x – 1)(x + 2) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y = {x^2} + 1\\\left[ \begin{array}{l}x = 1\\x =  – 2\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 1\\y = 2\,\,\,\left( {tm} \right)\end{array} \right.\\\left\{ \begin{array}{l}x =  – 2\\y = 5\,\,\,\left( {tm} \right)\end{array} \right.\end{array} \right.\end{array}\).