## Tính: $\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}$

Question

Tính: $\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}$

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1 year 2020-11-29T05:13:12+00:00 2 Answers 33 views 0

1. Đáp án:

$2 + \sqrt 2 – \sqrt 3 – \sqrt 6$

Giải thích các bước giải:

$\begin{array}{l} \dfrac{{2\sqrt {15} – 2\sqrt {10} + \sqrt 6 – 3}}{{2\sqrt 5 – 2\sqrt {10} – \sqrt 3 + \sqrt 6 }}\\ = \dfrac{{\left( {2\sqrt {15} – 3} \right) – \left( {2\sqrt {10} – \sqrt 6 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right) – \left( {2\sqrt {10} – \sqrt 6 } \right)}}\\ = \dfrac{{\sqrt 3 \left( {2\sqrt 5 – \sqrt 3 } \right) – \sqrt 2 \left( {2\sqrt 5 – \sqrt 3 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right) – \sqrt 2 \left( {2\sqrt 5 – \sqrt 3 } \right)}}\\ = \dfrac{{\left( {2\sqrt 5 – \sqrt 3 } \right)\left( {\sqrt 3 – \sqrt 2 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right)\left( {1 – \sqrt 2 } \right)}}\\ = \dfrac{{\sqrt 3 – \sqrt 2 }}{{1 – \sqrt 2 }}\\ = \dfrac{{\left( {\sqrt 3 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}{{\left( {1 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\ = – \left( {\sqrt 3 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)\\ = 2 + \sqrt 2 – \sqrt 3 – \sqrt 6 \end{array}$

2. Đáp án:

Giải thích các bước giải:

(2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3)/(2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6

=(2\sqrt{5}.\sqrt{3}-2\sqrt{5}.\sqrt{2}+\sqrt{2}.\sqrt{3}-\sqrt{3}.\sqrt{3})/(2\sqrt{5}-2\sqrt{5}.\sqrt{2}-\sqrt{3}+\sqrt{3}.\sqrt{2})

=[2\sqrt{5}(\sqrt{3}-\sqrt{2})-\sqrt{3}(\sqrt{3}-\sqrt{2})]/[2\sqrt{5}(1-\sqrt{2})-\sqrt{3}(1-\sqrt{2})

=[(2\sqrt{5}-3)(\sqrt{3}-\sqrt{2})]/[(2\sqrt{5}-\sqrt{3})(1-\sqrt{2})

=(\sqrt{3}-\sqrt{2})/(1-\sqrt{2