Question

cho x=\(\dfrac{\sqrt{3}}{4}\) tính A=\(\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}\)

Answer 1

Ta có: x = \(\dfrac{\sqrt{3}}{4}\)⇒2x=\(\dfrac{\sqrt{3}}{2}\)

⇒ 1-2x = 1 - \(\dfrac{\sqrt{3}}{2}\) = \(\left(\dfrac{\sqrt{3}-1}{2}\right)^2\)

TT ⇒ 1+ 2x = \(\left(\dfrac{\sqrt{3}+1}{2}\right)^2\)

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

= 1

Vây với x = \(\dfrac{\sqrt{3}}{4}\) thì A = 1