Question

bài 1:trục căn thức ở mẫu

a.\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)

b.\(\dfrac{37}{7+2\sqrt{3}}\)

c.\(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}\)với a>0, a≠4

d.\(\dfrac{1+\sqrt{a}}{2-\sqrt{a}}\)

Answer 1

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{5-2\sqrt{5}+1}{5-1}=\dfrac{2\left(3-\sqrt{5}\right)}{4}=\dfrac{3-\sqrt{5}}{2}\)

Answer 2

b: \(\dfrac{37}{7+2\sqrt{3}}=7-2\sqrt{3}\)

c:\(=\dfrac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{5}\right)}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\)

d: \(=\dfrac{\left(1+\sqrt{a}\right)\cdot\left(2+\sqrt{a}\right)}{4-a}\)