π
Đề :
$(\sqrt{6}+\sqrt{10})\sqrt{4-\sqrt{15}}$
Giải :
$(\sqrt{6}+\sqrt{10})\sqrt{4-\sqrt{15}}$
= $(\sqrt{10}+\sqrt{6})\sqrt{4-\sqrt{15}}$
= $(\sqrt{2.5}+\sqrt{2.3})\sqrt{4-\sqrt{5.3}}$
= $\sqrt{2}(\sqrt{5}+\sqrt{3})\sqrt{4-\sqrt{5.3}}$
= $(\sqrt{5}+\sqrt{3})\sqrt{2}\sqrt{4-\sqrt{5.3}}$
= $(\sqrt{5}+\sqrt{3})\sqrt{2(4-\sqrt{5.3})}$
= $(\sqrt{5}+\sqrt{3})\sqrt{8-2\sqrt{5.3}}$
= $(\sqrt{5}+\sqrt{3})\sqrt{5+3-2\sqrt{5.3}}$
= $(\sqrt{5}+\sqrt{3})\sqrt{\sqrt{5}^2+\sqrt{3}^2-2\sqrt{5.3}}$
= $(\sqrt{5}+\sqrt{3})\sqrt{\sqrt{5}^2-2\sqrt{5.3}+\sqrt{5}^2}$
= $(\sqrt{5}+\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}$
= $(\sqrt{5}+\sqrt{3})|\sqrt{5}-\sqrt{3}|$
= $(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})$
= $\sqrt{5}^2-\sqrt{3}^2$
= $5-3$
= $2$
$\text{Chúc bạn học tốt nhé :x ^_^ XD}$
