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Đáp án:
a) $\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=-2$
b) $\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=4$
Bài 5:
$A=-\sqrt{6}$
Giải thích các bước giải:
a)
$\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\=\dfrac{2(5+\sqrt{10})}{\sqrt{5}+\sqrt{2}}+\dfrac{8(1+\sqrt{5})}{(1-\sqrt{5})(1+\sqrt{5})}\\=\dfrac{2.\sqrt{5}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}+\dfrac{8(1+\sqrt{5})}{1-5}\\=2\sqrt{5}-2(1+\sqrt{5})\\=-2$
c)
$\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\=\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}+\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\\=\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\\=\dfrac{4}{\sqrt{4-3}}\\=4$
Bài 5:
$A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\\=\sqrt{3(4-\sqrt{7})}-\sqrt{3(4+\sqrt{7})}\\=\sqrt{3}\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{3}\sqrt{\dfrac{8+2\sqrt{7}}{2}}\\=\sqrt{3}\sqrt{\dfrac{\sqrt{(\sqrt{7}-1)^2}}{2}}-\sqrt{3}\sqrt{\dfrac{(\sqrt{7}+1)^2}{2}}\\=\sqrt{3}.\dfrac{|\sqrt{7}-1|}{\sqrt{2}}-\sqrt{3}.\dfrac{\sqrt{7}+1}{\sqrt{2}}\\=\sqrt{3}.\dfrac{\sqrt{7}-1}{\sqrt{2}}-\sqrt{3}.\dfrac{\sqrt{7}+1}{\sqrt{2}}\\=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\\=\dfrac{-2\sqrt{3}}{\sqrt{2}}\\=\dfrac{-2\sqrt{6}}{2}\\=-\sqrt{6}$
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