√14+6 √5 √5+2 √6- √5-2 √6 √9-4 √5+ √6+2 √5 √15-6 √6+ √10+4 √6- √7+2 √6 05/07/2021 Bởi Athena √14+6 √5 √5+2 √6- √5-2 √6 √9-4 √5+ √6+2 √5 √15-6 √6+ √10+4 √6- √7+2 √6
a)$\sqrt {14 + 6\sqrt 5 } = \sqrt {9 + 5 + 2.3.\sqrt 5 } = \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} = 3 + \sqrt 5 $ b) $\begin{array}{l} \sqrt {5 + 2\sqrt 6 } – \sqrt {5 – 2\sqrt 6 } \\ = \sqrt {\left( {2 + 3 + 2.\sqrt 2 .\sqrt 3 } \right)} – \sqrt {\left( {2 + 3 – 2.\sqrt 2 .\sqrt 3 } \right)} \\ = \sqrt {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}} \\ = \sqrt 2 + \sqrt 3 – \left( {\sqrt 3 – \sqrt 2 } \right) = 2\sqrt 2 \end{array}$ c) $\begin{array}{l} \sqrt {9 – 4\sqrt 5 } + \sqrt {6 + 2\sqrt 5 } \\ = \sqrt {\left( {5 + 4 – 2.2\sqrt 5 } \right)} + \sqrt {\left( {5 + 1 + 2.1.\sqrt 5 } \right)} \\ = \sqrt {{{\left( {\sqrt 5 – 2} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\ = \sqrt 5 – 2 + \sqrt 5 + 1 = 2\sqrt 5 – 1 \end{array}$ d) $\begin{array}{l} \sqrt {15 – 6\sqrt 6 } + \sqrt {10 + 4\sqrt 6 } – \sqrt {7 + 2\sqrt 6 } \\ = \sqrt {9 + 6 – 2.3.\sqrt 6 } + \sqrt {6 + 4 + 2.2.\sqrt 6 } – \sqrt {6 + 1 + 2.1.\sqrt 6 } \\ = \sqrt {{{\left( {3 – \sqrt 6 } \right)}^2}} + \sqrt {{{\left( {\sqrt 6 + 2} \right)}^2}} – \sqrt {{{\left( {\sqrt 6 + 1} \right)}^2}} \\ = 3 – \sqrt 6 + \sqrt 6 + 2 – \sqrt 6 – 1\\ = 4 – \sqrt 6 \end{array}$ Bình luận
a)$\sqrt {14 + 6\sqrt 5 } = \sqrt {9 + 5 + 2.3.\sqrt 5 } = \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} = 3 + \sqrt 5 $
b)
$\begin{array}{l} \sqrt {5 + 2\sqrt 6 } – \sqrt {5 – 2\sqrt 6 } \\ = \sqrt {\left( {2 + 3 + 2.\sqrt 2 .\sqrt 3 } \right)} – \sqrt {\left( {2 + 3 – 2.\sqrt 2 .\sqrt 3 } \right)} \\ = \sqrt {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}} \\ = \sqrt 2 + \sqrt 3 – \left( {\sqrt 3 – \sqrt 2 } \right) = 2\sqrt 2 \end{array}$
c)
$\begin{array}{l} \sqrt {9 – 4\sqrt 5 } + \sqrt {6 + 2\sqrt 5 } \\ = \sqrt {\left( {5 + 4 – 2.2\sqrt 5 } \right)} + \sqrt {\left( {5 + 1 + 2.1.\sqrt 5 } \right)} \\ = \sqrt {{{\left( {\sqrt 5 – 2} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\ = \sqrt 5 – 2 + \sqrt 5 + 1 = 2\sqrt 5 – 1 \end{array}$
d)
$\begin{array}{l} \sqrt {15 – 6\sqrt 6 } + \sqrt {10 + 4\sqrt 6 } – \sqrt {7 + 2\sqrt 6 } \\ = \sqrt {9 + 6 – 2.3.\sqrt 6 } + \sqrt {6 + 4 + 2.2.\sqrt 6 } – \sqrt {6 + 1 + 2.1.\sqrt 6 } \\ = \sqrt {{{\left( {3 – \sqrt 6 } \right)}^2}} + \sqrt {{{\left( {\sqrt 6 + 2} \right)}^2}} – \sqrt {{{\left( {\sqrt 6 + 1} \right)}^2}} \\ = 3 – \sqrt 6 + \sqrt 6 + 2 – \sqrt 6 – 1\\ = 4 – \sqrt 6 \end{array}$