Đáp án:
2) -1
Giải thích các bước giải:
\(\begin{array}{l}
2)\dfrac{{\sqrt {{{\left( {3 - x} \right)}^2}} }}{{x - 3}} = \dfrac{{\left| {3 - x} \right|}}{{x - 3}}\\
= \dfrac{{3 - x}}{{ - \left( {3 - x} \right)}}\left( {do:x < 3} \right)\\
= - 1\\
4)\dfrac{{\sqrt {{{\left( {3x - 1} \right)}^2}} }}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}} = \dfrac{{\left| {3x - 1} \right|}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{ - \left( {3x - 1} \right)}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}} = - \dfrac{1}{{3x + 1}}\\
6)\sqrt {{{\left( {2x - 1} \right)}^2}} - \sqrt {{{\left( {2x + 1} \right)}^2}} \\
= \left| {2x - 1} \right| - \left| {2x + 1} \right|\\
= - \left( {2x - 1} \right) - \left( { - 2x - 1} \right)\\
= - 2x + 1 + 2x + 1 = 2\\
8)N = \sqrt {x + \sqrt {2x - 1} } - \sqrt {x - \sqrt {2x - 1} } \\
= \dfrac{{\sqrt {2x + 2\sqrt {2x - 1} } - \sqrt {2x - 2\sqrt {2x - 1} } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {2x - 1 + 2\sqrt {2x - 1} .1 + 1} - \sqrt {2x - 1 - 2\sqrt {2x - 1} .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\left| {\sqrt {2x - 1} + 1} \right| - \left| {\sqrt {2x - 1} - 1} \right|}}{{\sqrt 2 }}\\
\to \left[ \begin{array}{l}
N = \dfrac{{\sqrt {2x - 1} + 1 - \left( {\sqrt {2x - 1} - 1} \right)}}{{\sqrt 2 }}\left( {DK:\sqrt {2x - 1} - 1 \ge 0 \to x \ge 1} \right)\\
N = \dfrac{{\sqrt {2x - 1} + 1 - \left( { - \sqrt {2x - 1} + 1} \right)}}{{\sqrt 2 }}\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
N = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 \\
N = \dfrac{{2\sqrt {2x - 1} }}{{\sqrt 2 }} = \sqrt {4x - 2}
\end{array} \right.
\end{array}\)
