Đáp án:
\[A = 2 - \sqrt 3 \]
Giải thích các bước giải:
*) Giair phương trình \(\sqrt {2{x^2} - 3x - 5} = x - 1\)
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge \frac{5}{2}\\
x \le - 1
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\sqrt {2{x^2} - 3x - 5} = x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
2{x^2} - 3x - 5 = {\left( {x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
2{x^2} - 3x - 5 = {x^2} - 2x + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
{x^2} - x - 6 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 3\left( {t/m} \right)\\
A = \frac{{\sqrt x - 1}}{{\sqrt x + 1}} = \frac{{\sqrt 3 - 1}}{{\left( {\sqrt 3 + 1} \right)}} = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} = \frac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3
\end{array}\)
