Giúp mình câu này ạ
$a. (3+$ $\sqrt[]{5})($$\sqrt[]{10}-2)($ $\sqrt[]{3-}$ $\sqrt[n]{5})$
$b. (4+$ $\sqrt[]{15})($ $\sqrt[]{10}-$ $\sqrt[]{6})$ $\sqrt[]{4-}$ $\sqrt[]{15}$
Giúp mình câu này ạ
$a. (3+$ $\sqrt[]{5})($$\sqrt[]{10}-2)($ $\sqrt[]{3-}$ $\sqrt[n]{5})$
$b. (4+$ $\sqrt[]{15})($ $\sqrt[]{10}-$ $\sqrt[]{6})$ $\sqrt[]{4-}$ $\sqrt[]{15}$
Đáp án:
$\begin{array}{l}
a)\left( {3 + \sqrt 5 } \right)\left( {\sqrt {10} – 2} \right)\left( {\sqrt {3 – \sqrt 5 } } \right)\\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 – \sqrt 2 } \right).\sqrt 2 .\sqrt {3 – \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 – \sqrt 2 } \right).\sqrt {6 – 2\sqrt 5 } \\
= \dfrac{1}{2}.\left( {6 + 2\sqrt 5 } \right).\left( {\sqrt 5 – \sqrt 2 } \right).\sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} \\
= \dfrac{1}{2}{\left( {\sqrt 5 + 1} \right)^2}.\left( {\sqrt 5 – 1} \right).\left( {\sqrt 5 – \sqrt 2 } \right)\\
= \dfrac{1}{2}.\left( {\sqrt 5 + 1} \right).\left( {5 – 1} \right).\left( {\sqrt 5 – 2} \right)\\
= 2.\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 – 2} \right)\\
= 6 – 2\sqrt 5 \\
b)\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} – \sqrt 6 } \right).\sqrt {4 – \sqrt {15} } \\
= \dfrac{1}{2}.\left( {8 + 2\sqrt {15} } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt 2 .\sqrt {4 – \sqrt {15} } \\
= \dfrac{1}{2}\left( {5 + 2\sqrt 5 .\sqrt 3 + 3} \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {8 – 2\sqrt {15} } \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} \\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.\left( {\sqrt 5 – \sqrt 3 } \right).\left( {\sqrt 5 – \sqrt 3 } \right)\\
= \dfrac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.{\left( {\sqrt 5 – \sqrt 3 } \right)^2}\\
= \dfrac{1}{2}.{\left( {5 – 3} \right)^2}\\
= 2
\end{array}$