Question

[tuananh vu]

\(\sqrt{x+2\sqrt{x-4}}+\sqrt{x-2\sqrt{2x-4}}\)với x lớn hơn hoặc bằng 2

Answer 1

[Le Nhat Phuong]

đk : x ≥ 2 
Bạn bình phương 2 vế, thu gọn đc: 
3√[x(x−2)(x+1)] ≤ 2x2−6x−2 
<=> 3√[(x2−2x)(x+1)] ≤ 2(x2−2x) − 2(x+1) 
Chia 2 vế cho (x+1), đặt t= căn((x2−2x)/(x+1)), t≥ 0 ta đc: 
2t^2 - 3t - 2 ≥ 0 => t ≥ 2 
<=> x^2 - 2x ≥ 4x + 4 
<=> x^2 - 6x -4 ≥ 0 
<=> x ≥ 3+√13

P/s: Tham khảo nhé

Answer 2

[0o0 o0o khùng]

\(\sqrt{x+2\sqrt{x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

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

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

\(=\sqrt{x+2.\left|\sqrt{x}-2\right|}+\sqrt{x-2.\left|\sqrt{2x}-2\right|}\)

\(=\sqrt{x+2.\left(\sqrt{x}-2\right)}+\sqrt{x-2.\left(\sqrt{2x}-2\right)}\)

\(=\sqrt{x+2\sqrt{x}-4}+\sqrt{x-2\sqrt{2x}+4}\)

\(=\left(\sqrt{x+2\sqrt{x}-4}\right)^2+\left(\sqrt{x-2\sqrt{2x}+4}\right)^2\)

\(=x+2\sqrt{x}-4+x-2\sqrt{2x}+4\)

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

\(=2x+2\sqrt{x}-2\sqrt{2}.\sqrt{x}\)

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