Đáp án:
\(\begin{array}{l}
a. - \sqrt 2 \\
b.\dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.3\sqrt {2x} + \sqrt {2x} = 4\sqrt {2x} \\
b.\dfrac{{a - b}}{{{b^2}}}.\dfrac{{\left| a \right|{b^2}}}{{\sqrt {{{\left( {a - b} \right)}^2}} }} = \dfrac{{a - b}}{{{b^2}}}.\dfrac{{a{b^2}}}{{\left| {a - b} \right|}}\\
= \dfrac{{a - b}}{{{b^2}}}.\dfrac{{a{b^2}}}{{a - b}}\left( {Do:a > b > 0} \right)\\
= a\\
d.\dfrac{{\left| x \right| + \sqrt {{{\left( {2 - x} \right)}^2}} + 1}}{{x - 1}} = \dfrac{{x + \left| {2 - x} \right| + 1}}{{x - 1}}\left( {do:x > 2} \right)\\
= \dfrac{{x - \left( {2 - x} \right) + 1}}{{x - 1}} = \dfrac{{2x - 1}}{{x - 1}}\\
a.\left( {\dfrac{{3.\sqrt 2 .\sqrt 3 }}{2} + \dfrac{{2\sqrt 2 }}{{\sqrt 3 }} - \dfrac{{4\sqrt 3 }}{{\sqrt 2 }}} \right)\left( {\dfrac{{3\sqrt 2 }}{{\sqrt 3 }} - 2\sqrt 3 - \sqrt 6 } \right)\\
= \left( {\dfrac{{3.\sqrt 3 }}{{\sqrt 2 }} + \dfrac{{2\sqrt 2 }}{{\sqrt 3 }} - \dfrac{{4\sqrt 3 }}{{\sqrt 2 }}} \right)\left( {\dfrac{{3\sqrt 2 }}{{\sqrt 3 }} - 2\sqrt 3 - \sqrt 6 } \right)\\
= \left( {\dfrac{{3.3}}{{\sqrt 6 }} + \dfrac{{2.2}}{{\sqrt 6 }} - \dfrac{{4.3}}{{\sqrt 6 }}} \right)\left( {\sqrt 6 - 2\sqrt 3 - \sqrt 6 } \right)\\
= \dfrac{1}{{\sqrt 6 }}.\left( { - 2\sqrt 3 } \right) = - \sqrt 2 \\
b.\left( {\dfrac{{\sqrt 7 + \sqrt 3 - \sqrt 7 + \sqrt 3 }}{{7 - 3}} + 1} \right).\dfrac{1}{{3 + 2\sqrt 3 + 1}}\\
= \left( {\dfrac{{2\sqrt 3 }}{4} + 1} \right).\dfrac{1}{{4 + 2\sqrt 3 }}\\
= \left( {\dfrac{{\sqrt 3 + 2}}{2}} \right).\dfrac{1}{{2\left( {2 + \sqrt 3 } \right)}}\\
= \dfrac{1}{4}
\end{array}\)
