Tính: $\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}$
Tính: $\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}$
By Anna
By Anna
Tính: $\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}$
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`
Đáp án:
$2 + \sqrt 2 – \sqrt 3 – \sqrt 6 $
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{2\sqrt {15} – 2\sqrt {10} + \sqrt 6 – 3}}{{2\sqrt 5 – 2\sqrt {10} – \sqrt 3 + \sqrt 6 }}\\
= \dfrac{{\left( {2\sqrt {15} – 3} \right) – \left( {2\sqrt {10} – \sqrt 6 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right) – \left( {2\sqrt {10} – \sqrt 6 } \right)}}\\
= \dfrac{{\sqrt 3 \left( {2\sqrt 5 – \sqrt 3 } \right) – \sqrt 2 \left( {2\sqrt 5 – \sqrt 3 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right) – \sqrt 2 \left( {2\sqrt 5 – \sqrt 3 } \right)}}\\
= \dfrac{{\left( {2\sqrt 5 – \sqrt 3 } \right)\left( {\sqrt 3 – \sqrt 2 } \right)}}{{\left( {2\sqrt 5 – \sqrt 3 } \right)\left( {1 – \sqrt 2 } \right)}}\\
= \dfrac{{\sqrt 3 – \sqrt 2 }}{{1 – \sqrt 2 }}\\
= \dfrac{{\left( {\sqrt 3 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}{{\left( {1 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\
= – \left( {\sqrt 3 – \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)\\
= 2 + \sqrt 2 – \sqrt 3 – \sqrt 6
\end{array}$