Question

[Phước Nhanh Nguyễn]

Cho \(x=\frac{\sqrt{28-16\sqrt{3}}}{\sqrt{3}-1}\).Tính giá trị của \(P=\left(x^2+2x-1\right)^{2012}\)

Answer 1

[Đỗ Tuấn Anh]

\(x=\frac{\sqrt{28-16\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.4+4^2}}{\sqrt{3}-1}\)\(=\)\(\frac{\sqrt{\left(2\sqrt{3}-4\right)^2}}{\sqrt{3}-1}\)\(=\frac{4-2\sqrt{3}}{\sqrt{3}-1}=\frac{3-2\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(\sqrt{3}-1\right)^2}{\sqrt{3}-1}=\sqrt{3}-1\)

Thay \(x=\sqrt{3}-1\)vào bieu thuc P, ta duoc

\(\left[\left(\sqrt{3}-1\right)^2+2\left(\sqrt{3}-1\right)-1\right]^{2012}\)\(=\left(4-2\sqrt{3}+2\sqrt{3}-2-1\right)^{2012}=1^{2012}=1\)

Vậy P=1