Đáp án:
c) \(0 < a < 1\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x - ay = a\\
ax + y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - ay = a\\
{a^2}x + ay = a
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {1 + {a^2}} \right)x = 2a\\
x - ay = a
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a}}{{{a^2} + 1}}\\
y = \dfrac{{x - a}}{a}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a}}{{{a^2} + 1}}\\
y = \dfrac{{\dfrac{{2a}}{{{a^2} + 1}} - a}}{a} = \dfrac{{2a - {a^3} - a}}{{a\left( {{a^2} + 1} \right)}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2a}}{{{a^2} + 1}}\\
y = \dfrac{{ - {a^2} + 1}}{{{a^2} + 1}}
\end{array} \right.\\
a)Thay:a = \sqrt 2 - 1\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2\left( {\sqrt 2 - 1} \right)}}{{{{\left( {\sqrt 2 - 1} \right)}^2} + 1}} = \dfrac{{\sqrt 2 }}{2}\\
y = \dfrac{{ - {{\left( {\sqrt 2 - 1} \right)}^2} + 1}}{{{{\left( {\sqrt 2 - 1} \right)}^2} + 1}} = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
b)Do:{a^2} + 1 > 0\forall a\\
\to dpcm\\
b)Do:x > 0;y > 0\\
\to \left\{ \begin{array}{l}
\dfrac{{2a}}{{{a^2} + 1}} > 0\\
\dfrac{{ - {a^2} + 1}}{{{a^2} + 1}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a > 0\\
\left( {1 - a} \right)\left( {1 + a} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
a > 0\\
- 1 < a < 1
\end{array} \right.\\
KL:0 < a < 1
\end{array}\)
