Question

Mềnh cần gấp ạ, giúp mềnh mềnh xin biếu 5coin mần quà :))

Answer 1

c) \(=\dfrac{\sqrt{5}\left(2-7\sqrt{3}\right)}{2-7\sqrt{3}}+\dfrac{31\left(6-\sqrt{5}\right)}{36-5}=\sqrt{5}+6-\sqrt{5}=6\)

d) \(=\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{10}}+\dfrac{10\sqrt{5}}{5}+\left|3\sqrt{5}-7\right|=\sqrt{5}+1+2\sqrt{5}+7-3\sqrt{5}=8\)

e) \(=\dfrac{12\left(4+\sqrt{10}\right)}{16-10}-\sqrt{\left(\sqrt{10}+2\right)^2}-\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\left(4+\sqrt{10}\right)-\sqrt{10}-2-\sqrt{10}=6\)

P/s: Nhớ lời hứa nha bé =))

Answer 2

c) \(\dfrac{2\sqrt{5}-7\sqrt{15}}{2-7\sqrt{3}}+\dfrac{31}{6+\sqrt{5}}=\dfrac{\sqrt{5}\left(2-7\sqrt{3}\right)}{2-7\sqrt{3}}+\dfrac{31}{6+\sqrt{5}}=\sqrt{5}+\dfrac{31}{6+\sqrt{5}}=\dfrac{\sqrt{5}\left(6+\sqrt{5}\right)+31}{6+\sqrt{5}}=\dfrac{6\sqrt{5}+36}{6+\sqrt{5}}=\dfrac{6\left(\sqrt{5}+6\right)}{6+\sqrt{5}}=6\)

d) \(\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{10}}+\dfrac{10}{\sqrt{5}}+\sqrt{\left(3\sqrt{5}-7\right)^2}=\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{10}}+\dfrac{\sqrt{100}}{\sqrt{5}}+\sqrt{\left(7-3\sqrt{5}\right)^2}=\sqrt{5}+1+2\sqrt{5}+7-3\sqrt{5}=8\)

e) \(\dfrac{12}{4-\sqrt{10}}-\sqrt{14+4\sqrt{10}}-\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}=\dfrac{\sqrt{144}}{\sqrt{16}-\sqrt{10}}-\sqrt{14+2\sqrt{40}}-\dfrac{\sqrt{50}+\sqrt{10}}{\sqrt{5}+1}=\dfrac{\sqrt{144}}{\sqrt{2}\left(\sqrt{8}-\sqrt{5}\right)}-\sqrt{10+2\sqrt{10}.\sqrt{4}+4}-\dfrac{\sqrt{10}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=\dfrac{\sqrt{72}}{\sqrt{8}-\sqrt{5}}-\sqrt{10}-2-\sqrt{10}=\dfrac{\sqrt{72}}{\sqrt{8}-\sqrt{5}}-2\sqrt{10}-2=6\)