Thực hiện các phép tính sau
3.
c) $\sqrt{8\sqrt 3}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}=2\sqrt{2\sqrt 3}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0$
e) $\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{6-2\sqrt 5}+\sqrt{6+2\sqrt 5}}{\sqrt 2}=\dfrac{\sqrt{(\sqrt{5}-1)^2}+\sqrt{(\sqrt{5}+1)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}$
4.
a) $A=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\sqrt{3}$
b) $\dfrac{1}{1-\sqrt{2}}+\dfrac{1}{1+\sqrt{2}}=\dfrac{1+\sqrt{2}+1-\sqrt{2}}{1-2}=-2$
c) $\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{(5+\sqrt{5})^2+(5-\sqrt{5})^2}{25-5}=\dfrac{60}{20}=3$
d) $\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}=\dfrac{\sqrt{3}(\sqrt{\sqrt{3}+1}+1)-\sqrt{3}(\sqrt{\sqrt{3}+1}-1)}{\sqrt{3}+1-1}=\dfrac{2\sqrt{3}}{\sqrt{3}}=2$
$25-5=10$ à bạn? ^^