Question

Cho A = \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\) với x \(\ge\) 0

tính GT A khi x = \(\frac{1}{2}\left(\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\right)\)

Answer 1

ta có\(x=\frac{1}{2}\left(\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\right)\)

⇔\(x=\frac{1}{2}\left(\sqrt{2+4+4\sqrt{2}}+\sqrt{2+4-4\sqrt{2}}\right)\)(hằng dẳng thức)

⇔\(x=\frac{1}{2}\left(\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\right)\)

⇔\(x=\frac{1}{2}\left(\left|2+\sqrt{2}\right|+\left|2-\sqrt{2}\right|\right)\)

⇔\(x=\frac{1}{2}\left(2+\sqrt{2}+2-\sqrt{2}\right)\)

⇔\(x=\frac{1}{2}\cdot4\)

⇔x=2

thay x=2 vào biểu thức ta có:

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

=\(2-2\sqrt{2}+1\)

=\(3-2\sqrt{2}\)