a)=\(3\sqrt{5}-8\sqrt{5}+6\sqrt{2}=-5\sqrt{5}+6\sqrt{2}\)
c)=\(\sqrt{8+2\sqrt{6-2\sqrt{5}}}=\sqrt{8+2\sqrt{\left(\sqrt{5}-1\right)^2}}\)
=\(\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{8+2\sqrt{5}-2}\)
=\(\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
e)=\(\dfrac{4\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\dfrac{3\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\dfrac{16\left(\sqrt{5}+3\right)}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}\)
=\(\dfrac{4\sqrt{5}+4}{4}+\dfrac{3\sqrt{5}+6}{1}+\dfrac{16\sqrt{5}+48}{-4}\)
=\(\sqrt{5}+1+3\sqrt{5}+6-4\sqrt{5}-12\)=-5
BÀi 1
a)\(=3\sqrt{5}-8\sqrt{5}+6\sqrt{2}=6\sqrt{2}-5\sqrt{5}\)
c)\(=\sqrt{8+2\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{6+2\sqrt{5}}=\sqrt{\left(1+\sqrt{5}\right)^2}=1+\sqrt{5}\)
e)\(=\dfrac{4\left(\sqrt{5}+1\right)}{4}+\dfrac{3\left(\sqrt{5}+2\right)}{3}+\dfrac{16\left(\sqrt{5}+3\right)}{2}=\sqrt{5}+1+\sqrt{5}+2+8\sqrt{5}+24=10\sqrt{5}+27\)
