Đáp án:
1b) -3
Giải thích các bước giải:
\(\begin{array}{l}
1b)\left[ {\dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 4 } \right)}}{{\sqrt 5 - 2}} - \dfrac{{2\sqrt 3 \left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }}} \right].\sqrt 3 \\
= \left[ {\dfrac{{\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 - 2}} - \dfrac{{2\sqrt 3 \left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }}} \right].\sqrt 3 \\
= \left( {\sqrt 3 - 2\sqrt 3 } \right).\sqrt 3 \\
= - \sqrt 3 .\sqrt 3 = - 3
\end{array}\)
\(\begin{array}{l}
2b)\left( {\dfrac{{4\sqrt {10} - 4\sqrt 5 }}{{2\sqrt 2 - 2}} - \dfrac{{2\sqrt {10} - \sqrt {15} }}{{2\sqrt 2 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 5 }}\\
= \left[ {\dfrac{{4\sqrt 5 \left( {\sqrt 2 - 1} \right)}}{{2\left( {\sqrt 2 - 1} \right)}} - \dfrac{{\sqrt 5 \left( {2\sqrt 2 - \sqrt 3 } \right)}}{{2\sqrt 2 - \sqrt 3 }}} \right].\sqrt 5 \\
= \left( {2\sqrt 5 - \sqrt 5 } \right).\sqrt 5 \\
= \sqrt 5 .\sqrt 5 = 5
\end{array}\)
