Cho hệ phương trình : x+ay=1 và -ax+y=a . Chứng minh rằng hệ pt luôn luôn có nghiệm duy nhất vs mọi a . Tìm giá trị của a để hệ phương trình có nghiệm (x;y) sao cho x<1 ;y <1

Đáp án:

\(a \ne 0;\,\,a \ne 1\)

Giải thích các bước giải:

\(\eqalign{ & \left\{ \matrix{ x + ay = 1 \hfill \cr - ax + y = a \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = 1 - ay \hfill \cr - a\left( {1 - ay} \right) + y = a \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x = 1 - ay \hfill \cr - a + {a^2}y + y = a \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = 1 - ay \hfill \cr y\left( {{a^2} + 1} \right) = 2a \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x = 1 - ay \hfill \cr y = {{2a} \over {{a^2} + 1}}\,\,\left( {Do\,{a^2} + 1 > 0\,\,\forall a} \right) \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x = 1 - {{2{a^2}} \over {{a^2} + 1}} \hfill \cr y = {{2a} \over {{a^2} + 1}} \hfill \cr} \right. \cr & Theo\,\,bai\,\,ra\,\,ta\,\,co: \cr & \left\{ \matrix{ x < 1 \hfill \cr y < 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 1 - {{2{a^2}} \over {{a^2} + 1}} < 1 \hfill \cr {{2a} \over {{a^2} + 1}} < 1 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {{2{a^2}} \over {{a^2} + 1}} > 0 \hfill \cr 2a < {a^2} + 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ {a^2} > 0 \hfill \cr {a^2} - 2a + 1 > 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ a \ne 0 \hfill \cr {\left( {a - 1} \right)^2} > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ a \ne 0 \hfill \cr a \ne 1 \hfill \cr} \right. \cr & Vay\,\,a \ne 0;\,\,a \ne 1 \cr} \)

\[\begin{array}{l} \left\{ \begin{array}{l} x + ay = 1\\ - ax + y = a \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + ay = 1\,\,\,\,\,\left( 1 \right)\\ y = a + ax\,\,\,\,\left( 2 \right) \end{array} \right.\\ \Rightarrow \left( 1 \right) \Leftrightarrow x + a\left( {a + ax} \right) = 1\\ \Leftrightarrow x + {a^2} + {a^2}x = 1\\ \Leftrightarrow \left( {{a^2} + 1} \right)x = 1 - {a^2}\,\,\,\left( {\,*} \right)\\ Vi\,\,{a^2} + 1 > 0\,\,\,\,\,\forall a.\\ \Rightarrow \left( * \right)\,\,\,luon\,\,\,co\,\,\,nghiem\,\,\,\,duy\,\,\,nhat\,\,\,x = \frac{{1 - {a^2}}}{{{a^2} + 1}}\,\,\,\,voi\,\,\,moi\,\,a.\\ \Rightarrow Hpt\,\,luon\,\,\,co\,\,\,nghiem\,\,\,duy\,\,\,nhat\,\,\,voi\,\,\,moi\,\,a.\\ \Rightarrow y = a + ax = a + a.\frac{{1 - {a^2}}}{{{a^2} + 1}} = \frac{{a\left( {{a^2} + 1} \right) + a - {a^3}}}{{{a^2} + 1}} = \frac{{2a}}{{{a^2} + 1}}.\\ \Rightarrow Hpt\,\,\,co\,\,\,nghiem\,\,duy\,\,\,nhat\,\,\,\left( {x;\,\,y} \right) = \left( {\frac{{1 - {a^2}}}{{{a^2} + 1}};\,\,\frac{{2a}}{{{a^2} + 1}}} \right).\\ Theo\,\,de\,\,bai\,\,co:\,\,\,\left\{ \begin{array}{l} x < 1\\ y < 1 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \frac{{1 - {a^2}}}{{{a^2} + 1}} < 1\\ \frac{{2a}}{{{a^2} + 1}} < 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 1 - {a^2} < {a^2} + 1\\ 2a < {a^2} + 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 2{a^2} > 0\\ {a^2} - 2a + 1 > 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} a \ne 0\\ {\left( {a - 1} \right)^2} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a \ne 0\\ a \ne 1 \end{array} \right..\\ Vay\,\,a \ne 0,\,\,a \ne 1\,\,\,thoa\,\,\,man\,\,\,bai\,\,toan. \end{array}\]