Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
7 - 4\sqrt 3 = 4 - 2.2.\sqrt 3 + 3 = {\left( {2 - \sqrt 3 } \right)^2}\\
P < 7 - 4\sqrt 3 \\
\Leftrightarrow P < {\left( {2 - \sqrt 3 } \right)^2}\\
P = {\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow 0 \le P < 2 - \sqrt 3 \\
\Leftrightarrow 0 \le {\left( {x - 1} \right)^2} < 2 - \sqrt 3 \\
\Leftrightarrow - \sqrt {2 - \sqrt 3 } < x - 1 < \sqrt {2 - \sqrt 3 } \\
\Leftrightarrow - \sqrt {\dfrac{1}{2}.\left( {4 - 2\sqrt 3 } \right)} < x - 1 < \sqrt {\dfrac{1}{2}.\left( {4 - 2\sqrt 3 } \right)} \\
\Leftrightarrow - \sqrt {\dfrac{1}{2}{{\left( {\sqrt 3 - 1} \right)}^2}} < x - 1 < \sqrt {\dfrac{1}{2}.{{\left( {\sqrt 3 - 1} \right)}^2}} \\
\Leftrightarrow - \dfrac{{\sqrt 3 - 1}}{{\sqrt 2 }} < x - 1 < \dfrac{{\sqrt 3 - 1}}{{\sqrt 2 }}\\
\Leftrightarrow \dfrac{{ - \sqrt 3 + 1 + \sqrt 2 }}{{\sqrt 2 }} < x < \dfrac{{\sqrt 3 - 1 + \sqrt 2 }}{{\sqrt 2 }}
\end{array}\)
