Question

[Chanoppa]

Rút gọn

\(\sqrt{2x+\sqrt{4x+1}}\) + \(\sqrt{2x-\sqrt{4x-1}}\) ( với \(\frac{1}{4}\) <x < \(\frac{1}{2}\))

\(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

Answer 1

[Nguyễn Linh Chi]

Câu 1: Sửa lạ đề chút nhé : 4x + 1  -> 4x -1 

 Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)

=>  \(\sqrt{2}.A\)= ​\(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)​

= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)

= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)

Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)

nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2

=> \(A=2:\sqrt{2}=\sqrt{2}\)

Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)

=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)

=> ​\(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)​

=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)

=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)

=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)

\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)