(4+$\sqrt[2]{15}$ )*($\sqrt[2]{10}$ – $\sqrt[2]{6}$ )*$\sqrt[2]{(4-\sqrt[2]{15} }$ giúp mình với 12/11/2021 Bởi Kinsley (4+$\sqrt[2]{15}$ )*($\sqrt[2]{10}$ – $\sqrt[2]{6}$ )*$\sqrt[2]{(4-\sqrt[2]{15} }$ giúp mình với
Đáp án: 2 Giải thích các bước giải: $\begin{array}{l}\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} – \sqrt 6 } \right)\sqrt {4 – \sqrt {15} } \\ = \frac{1}{2}.2.\left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt 2 .\sqrt {4 – \sqrt {15} } \\ = \frac{1}{2}.\left( {8 + 2\sqrt {15} } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {8 – 2\sqrt {15} } \\ = \frac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} \\ = \frac{1}{2}.\left( {\sqrt 5 + \sqrt 3 } \right).\left( {\sqrt 5 + \sqrt 3 } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\left( {\sqrt 5 – \sqrt 3 } \right)\\ = \frac{1}{2}.\left( {5 – 3} \right).\left( {5 – 3} \right)\\ = \frac{1}{2}.2.2\\ = 2\end{array}$ Bình luận
Đáp án: 2
Giải thích các bước giải:
$\begin{array}{l}
\left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} – \sqrt 6 } \right)\sqrt {4 – \sqrt {15} } \\
= \frac{1}{2}.2.\left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt 2 .\sqrt {4 – \sqrt {15} } \\
= \frac{1}{2}.\left( {8 + 2\sqrt {15} } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {8 – 2\sqrt {15} } \\
= \frac{1}{2}.{\left( {\sqrt 5 + \sqrt 3 } \right)^2}.\left( {\sqrt 5 – \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} \\
= \frac{1}{2}.\left( {\sqrt 5 + \sqrt 3 } \right).\left( {\sqrt 5 + \sqrt 3 } \right).\left( {\sqrt 5 – \sqrt 3 } \right).\left( {\sqrt 5 – \sqrt 3 } \right)\\
= \frac{1}{2}.\left( {5 – 3} \right).\left( {5 – 3} \right)\\
= \frac{1}{2}.2.2\\
= 2
\end{array}$