Câu hỏi của Đừng gọi tôi là Jung Hae Ri

Question

Rút gọn biểu thức:

$a,\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}$

$b,\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)$

Nguyễn Việt Lâm · Accepted Answer

$A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}$

$A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}$

$A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2$

$A=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)$

$A=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2$

$B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)$

$B=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)$

$B=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)$

$B=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)=2$Câu trả lời: $A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}$