Question

trục căn thức

a) \(\dfrac{1}{\sqrt{x-1}};\dfrac{a+2}{\sqrt{a^2-4}};\dfrac{x-y}{\sqrt{x^2-y^2}};\dfrac{a}{\sqrt{x^2}}\) (n lẻ)

b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}\)

c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}\)

Answer 1

Answer 2

a) \(\dfrac{1}{\sqrt{x-1}}=\dfrac{\sqrt{x-1}}{x-1}\)

 \(\dfrac{a+2}{\sqrt{a^2-4}}=\dfrac{\sqrt{a+2}}{\sqrt{a-2}}=\dfrac{\sqrt{a^2-4}}{a-2}\)

\(\dfrac{x-y}{\sqrt{x^2-y^2}}=\dfrac{x-y}{\sqrt{\left(x-y\right)\left(x+y\right)}}=\dfrac{\sqrt{x-y}}{\sqrt{x+y}}=\dfrac{\sqrt{x^2-y^2}}{x+y}\)

\(\dfrac{a}{\sqrt{x^2}}=\dfrac{a}{\left|x\right|}\)

b) \(\dfrac{\sqrt{x^2-1}+1}{\sqrt{x^2-1}-1}=\dfrac{\left(\sqrt{x^2-1}+1\right)^2}{x^2-2}\)

c) \(\dfrac{2}{\sqrt{7-2\sqrt{6}}}=\dfrac{2}{\sqrt{6}-1}=\dfrac{2\left(\sqrt{6}+1\right)}{5}\)