Đáp án:
c. \(\left[ \begin{array}{l}
x > 4\\
x < - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \left\{ { - 1;3} \right\}\\
G = \frac{{\left( {2x - 3} \right)\left( {{x^2} - 2x + 1 - 4} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 3} \right)}}\\
= \frac{{\left( {2x - 3} \right)\left( {{x^2} - 2x - 3} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 3} \right)}}\\
= \frac{{\left( {2x - 3} \right)\left( {x - 3} \right)\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}\left( {x - 3} \right)}}\\
= \frac{{2x - 3}}{{x + 1}}\\
b.x = \sqrt {3 - 2\sqrt 2 } = \sqrt {{{\left( {\sqrt 2 } \right)}^2} - 2.\sqrt 2 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \sqrt 2 - 1\left( {do:\sqrt 2 > 1} \right)\\
Thay:x = \sqrt {3 - 2\sqrt 2 } \\
\to G = \frac{{2\left( {\sqrt 2 - 1} \right) - 3}}{{\sqrt 2 - 1 + 1}}\\
= \frac{{2\sqrt 2 - 5}}{{\sqrt 2 }}\\
c.G > 1\\
\to \frac{{2x - 3}}{{x + 1}} > 1\\
\to \frac{{2x - 3 - x - 1}}{{x + 1}} > 0\\
\to \frac{{x - 4}}{{x + 1}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 4 > 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 < 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 4\\
x < - 1
\end{array} \right.
\end{array}\)
