Question

(\(\sqrt{28}-2\sqrt{14}+\sqrt{7}\) )\(\sqrt{7}+7\sqrt{8}\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)-\(\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

Giải hộ giúp mik trong hôm nay nha mọi người!:))Cảm ơn nhiều!:)))

Answer 1

a. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+14\sqrt{2}=14-14\sqrt{2}+7+14\sqrt{2}=21\)

b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}-\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)

c. \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)