Question

[anh_tuấn_bùi]

trục căn thức ở mẫu 

\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)

Answer 1

[💋Bevis💋]

\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)

\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{\sqrt{3}+1}{\sqrt{3}^2-1^2}-\frac{\sqrt{3}-1}{\sqrt{3}^2-1^2}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{3}^2-1^2}\)

\(=\frac{2}{3-1}=\frac{2}{2}=1\)

Answer 2

[Nguyễn Hoàng Sơn]

Quy đồng lên ta có:
\(\frac{\sqrt{3}+1-\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

Áp dụng hằng đẳng thức ta có

\(\frac{2}{\left(\sqrt{3}\right)^2-1^2}=\frac{2}{3-1}=\frac{2}{2}=1\)