trục căn thức ở mẫu :
a,\(\frac{3}{\sqrt{5}};\frac{2\sqrt{3}}{\sqrt{2}};\frac{a}{\sqrt{b}};\frac{x+1}{\sqrt{x^2-1}}\)
b,\(\frac{1}{\sqrt{3}+\sqrt{2}};\frac{2}{2-\sqrt{3}};\frac{\sqrt{2}+1}{\sqrt{2}-1};\frac{3\sqrt{2}}{\sqrt{3}+1}\)
c,\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
d,\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{3}}}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
câu c chắc là như này
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}=1+\frac{1}{\sqrt{2}+\sqrt{3}}\) = \(1+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(1+\frac{\sqrt{2}-\sqrt{3}}{2-3}=1+\frac{\sqrt{2}-\sqrt{3}}{-1}\) = \(1-\sqrt{2}+\sqrt{3}\)
