## Cần ngay bây h ah .đề bài là tính gtri biểu thuc

Question

Cần ngay bây h ah .đề bài là tính gtri biểu thuc

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9 months 2020-10-31T04:10:00+00:00 1 Answers 61 views 0

$\begin{array}{l} a)\sqrt {5 – 2\sqrt 6 } + \sqrt {12} + \dfrac{1}{2}\sqrt 8 \\ = \sqrt {3 – 2\sqrt 3 .\sqrt 2 + 2} + 2\sqrt 3 + \dfrac{1}{2}.2\sqrt 2 \\ = \sqrt {{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}} + 2\sqrt 3 + \sqrt 2 \\ = \sqrt 3 – \sqrt 2 + 2\sqrt 3 + \sqrt 2 \\ = 3\sqrt 3 \\ b)\dfrac{{3\sqrt 2 – 6}}{{\sqrt 2 – 1}} + \dfrac{{6\sqrt 2 – 4}}{{\sqrt 2 – 3}}\\ = \dfrac{{3\sqrt 2 \left( {1 – \sqrt 2 } \right)}}{{\sqrt 2 – 1}} + \dfrac{{2\sqrt 2 \left( {3 – \sqrt 2 } \right)}}{{\sqrt 2 – 3}}\\ = – 3\sqrt 2 – 2\sqrt 2 \\ = – 5\sqrt 5 \\ c)\dfrac{{\sqrt 6 – \sqrt 8 }}{{\sqrt 3 – 2}} – \sqrt {3 – 2\sqrt 2 } + \dfrac{1}{{1 + \sqrt 2 }}\\ = \dfrac{{\sqrt 2 \left( {\sqrt 3 – 2} \right)}}{{\sqrt 3 – 2}} – \sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} + \dfrac{{\sqrt 2 – 1}}{{2 – 1}}\\ = \sqrt 2 – \left( {\sqrt 2 – 1} \right) + \sqrt 2 – 1\\ = \sqrt 2 \end{array}$