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Question

Rút gọn biểu thức:

$a,\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}$

$b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$

Nguyễn Việt Lâm · Accepted Answer

$A=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}$

$A=\sqrt{6-2\sqrt{4+\sqrt{12}}}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}$

$A=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1$

$B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}$

$B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}$

$B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}=\sqrt{5\sqrt{3}+25-5\sqrt{3}}$

$B=\sqrt{25}=5$Câu trả lời: $A=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}$

$A=\sqrt{6-2\sqrt{4+\sqrt{12}}}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}$

$A=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1$

$B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}$

$B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}$

$B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}=\sqrt{5\sqrt{3}+25-5\sqrt{3}}$

$B=\sqrt{25}=5$

Violympic toán 9