Tính A = (1 + căn 3/2) / (1 + (căn 1 + căn 3/2))

Đề bài: Tính $A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}$ .

Trả lời

Hướng dẫn giải:

$$\begin{gathered} A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}} \\ =\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{4+2\sqrt{3}}{4}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}} \\ =\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{{\left(\sqrt{3}+1\right)}^2}{4}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{{\left(\sqrt{3}-1\right)}^2}{4}}} \\ =\frac{\frac{2+\sqrt{3}}{2}}{1+\frac{\sqrt{3}+1}{2}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\frac{\sqrt{3}-1}{2}} \\ =\frac{\frac{2+\sqrt{3}}{2}}{\frac{3+\sqrt{3}}{2}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{3-\sqrt{3}}{2}} \\ =\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}} \\ =\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)} \\ =\frac{\left(6+\sqrt{3}-3\right)+\left(6-\sqrt{3}-3\right)}{9-3} \end{gathered}$$

$=\frac{6}{6}=1$

Vậy A = 1.