Chương I - Căn bậc hai. Căn bậc ba

NT

Tính:
1)a)\(2\sqrt{40.\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}\)

b)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)

d)\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

e) (\(\sqrt{\dfrac{1}{7}}-\sqrt[]{\dfrac{16}{7}}+\sqrt[]{\dfrac{9}{7}}\)):\(\sqrt{7}\)

2)

a)A=\(\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt[]{5+2\sqrt[]{3}}}\)

b) B=\(\sqrt{4+\sqrt{8}}\) . \(\sqrt{2+\sqrt{2+\sqrt{2}}}\) .\(\sqrt{2-\sqrt{2+\sqrt{2}}}\)

PL
12 tháng 7 2018 lúc 17:37

\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)

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