Câu hỏi của Trần Thị Ngân

Question

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

ngonhuminh · Accepted Answer

$A=\sqrt{\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}.\left(\sqrt{10}+\sqrt{6}\right)$

$A=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}.\left(\sqrt{2.5}+\sqrt{2.3}\right)$

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

$A=\sqrt{8+2\sqrt{3.5}}.\left(\sqrt{5}+\sqrt{3}\right)\ $

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

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

$A=8+2\sqrt{15}$Câu trả lời: $A=\sqrt{\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}.\left(\sqrt{10}+\sqrt{6}\right)$

$A=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}.\left(\sqrt{2.5}+\sqrt{2.3}\right)$

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

$A=\sqrt{8+2\sqrt{3.5}}.\left(\sqrt{5}+\sqrt{3}\right)\ $

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

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

$A=8+2\sqrt{15}$

Ôn tập chương 1: Căn bậc hai. Căn bậc ba