Tính: a)$\frac{6-3\sqrt{6}}{6}$ b)$\frac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}$ c) $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}$ d) (4+$\sqrt{5}$ )(

Tính:
a)$\frac{6-3\sqrt{6}}{6}$
b)$\frac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}$
c) $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}$
d) (4+$\sqrt{5}$ )($\sqrt{10}$-$\sqrt{6}$)$\sqrt{4-\sqrt{15}}$

0 bình luận về “Tính: a)$\frac{6-3\sqrt{6}}{6}$ b)$\frac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}$ c) $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}$ d) (4+$\sqrt{5}$ )(”

  1. a. $\dfrac{6 – 3\sqrt{6}}{6} = \dfrac{\sqrt{6}(\sqrt{6} – 3}{(\sqrt{6})^2} = \dfrac{\sqrt{3}(\sqrt{2} – \sqrt{3})}{\sqrt{2}.\sqrt{3}} = \dfrac{\sqrt{2} – \sqrt{3}}{\sqrt{2}}$ 

    b. $\dfrac{\sqrt{42} – 6}{\sqrt{21} – \sqrt{18}} = \dfrac{\sqrt{6}(\sqrt{7} – \sqrt{6})}{\sqrt{3}(\sqrt{7} – \sqrt{7})} = \dfrac{\sqrt{2}.\sqrt{3}}{\sqrt{3}} = \sqrt{2}$ 

    c. $\sqrt{\sqrt{5} – \sqrt{3 – \sqrt{29 – 6\sqrt{20}}}}$ 

    $= \sqrt{\sqrt{5} – \sqrt{3 – \sqrt{(2\sqrt{5} – 3)^2}}}$ 

    $= \sqrt{\sqrt{5} – \sqrt{3 – (2\sqrt{5} – 3)}}$

    $= \sqrt{\sqrt{5} – \sqrt{6 – 2\sqrt{5}}}$
    $= \sqrt{\sqrt{5} – \sqrt{(\sqrt{5} – 1)^2}}$ $= \sqrt{\sqrt{5} – \sqrt{5} + 1} = \sqrt{1} = 1$ 

    d. $(4 + \sqrt{15})(\sqrt{10} – \sqrt{6})\sqrt{4 – \sqrt{15}}$
    $= (4 + \sqrt{15}).\sqrt{2}(\sqrt{5} – \sqrt{3})\sqrt{4 – \sqrt{15}}$ 

    $= (4 + \sqrt{15})(\sqrt{5} – \sqrt{3})\sqrt{2(4 – \sqrt{15})}$ 

    $= (4 + \sqrt{15})(\sqrt{5} – \sqrt{3})\sqrt{(\sqrt{5} – \sqrt{3})^2}$ 

    $= (4 + \sqrt{15})(\sqrt{5} – \sqrt{3})(\sqrt{5} – \sqrt{3})$
    $= (4 + \sqrt{15})(5 + 3 – 2\sqrt{15}) = (4 +  \sqrt{15}).2.(4 – \sqrt{15}) = 2.(16 – 15) = 2$ 

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    (Câu d có sửa đề chút nhé!)

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