\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{1}=1\)

b,c

\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)

=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)

\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)

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

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

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ĐKXĐ: \(5\le x\le13\)

Đặt \(\sqrt{x-5}=t\ge0\Rightarrow x=t^2+5\)

Phương trình trở thành:

\(\sqrt{2+t}=\sqrt{13-\left(t^2+5\right)}\)

\(\Leftrightarrow\sqrt{t+2}=\sqrt{8-t^2}\)

\(\Leftrightarrow t+2=8-t^2\)

\(\Leftrightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-5}=2\)

\(\Rightarrow x=9\)

Gọi A= \(\sqrt{5-\sqrt{13+2\sqrt{11}}}\) - \(\sqrt{5+\sqrt{13+2\sqrt{11}}}\) 

Lấy A bình phương rồi áp dụng hằng đẳng thức số 2 sẽ ra:

A^2 = \(10-\) \(2\sqrt{25-\left(13+2\sqrt{11}\right)}\)

= \(10-2\sqrt{11-2\sqrt{11}+1}\)

= \(10-2\sqrt{\left(\sqrt{11}-1\right)^2}\)

= \(12-2\sqrt{11}\)

=\(11-2\sqrt{11}+1\)

= \(\left(\sqrt{11}-1\right)^2\)

Suy ra A= \(\sqrt{11}-1\)

\(a=\sqrt{5-\sqrt{13+2\sqrt{11}}}\); \(b=\sqrt{5+\sqrt{13+2\sqrt{11}}}\)dễ thấy \(a< b\)

ta có \(a^2+b^2=10;a.b=\left(\sqrt{11}-1\right)^{ }\).

Từ đây ta có \(\left(a-b\right)^2=\left(\sqrt{11}-1\right)^2\)kết hợp với a<b => a-b=1-\(\sqrt{11}\)

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)

\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{5}-1}\)

\(=\sqrt{4-\sqrt{5}}\)

c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(=3-2=1\)

d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)

\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3+1}\)

\(=2\sqrt{3}\)

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

==================================================

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

===========================================================

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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a) Ta có: \(\sqrt{2}\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)\)

\(=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}-1-\sqrt{5}-1=-2\)

b) Ta có: \(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{2}+2\sqrt{2}+1}\)

\(=\sqrt{14+32\sqrt{2}}\)

c) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{13+\sqrt{48}}}\)

\(=\sqrt{6+2\sqrt{5}-2\sqrt{3}-1}\)

\(=\sqrt{5+2\sqrt{5}-2\sqrt{3}}\)

Ta có:

\(\frac{1}{n\sqrt{n+4}+\left(n+4\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+4\right)}\left(\sqrt{n}+\sqrt{n+4}\right)}\)

\(=\frac{\sqrt{n+4}-\sqrt{n}}{4\sqrt{n\left(n+4\right)}}=\frac{1}{4}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+4}}\right)\)

Áp dụng vào bài toán ta được

\(\frac{1}{1\sqrt{5}+5\sqrt{1}}+\frac{1}{5\sqrt{9}+9\sqrt{5}}+...+\frac{1}{2009\sqrt{2013}+2013\sqrt{2009}}\)

\(=\frac{1}{4}.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{2009}}-\frac{1}{\sqrt{2013}}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{\sqrt{2013}}\right)\)