\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

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

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)

\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)

\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)

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

b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)

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

=10

d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)

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

\(=2\sqrt{5}+4\sqrt{2}\)

\(a,=\sqrt{3}+4\sqrt{3}+20\sqrt{3}-10\sqrt{3}=15\sqrt{3}\\ b,=4\sqrt{5}+\sqrt{5}-1-\dfrac{20\left(\sqrt{5}-1\right)}{4}\\ =5\sqrt{5}-1-5\sqrt{5}+5=4\\ c,=\dfrac{6\sqrt{13}+6+6\sqrt{13}-6}{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}=\dfrac{12\sqrt{13}}{12}=\sqrt{13}\\ d,=\left(\sin^238^0+\cos^238^0\right)+\left(\tan67^0-\tan67^0\right)=1+0=1\)

a: \(=\sqrt{3}+4\sqrt{3}+4\cdot5\sqrt{3}-10\sqrt{3}\)

\(=15\sqrt{3}\)

b: \(=2\cdot2\sqrt{5}+\sqrt{5}-1-5+5\sqrt{5}\)

=-6

\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)

\(\Leftrightarrow x=\sqrt{5+\sqrt{13+x}}\) (\(x\ge0\))

\(\Leftrightarrow x^2=5+\sqrt{13+x}\)

\(\Leftrightarrow x^2-9=\sqrt{13+x}-4\)

\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=\dfrac{x-3}{\sqrt{13+x}+4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=\dfrac{1}{\sqrt{x+13}+4}\left(∗\right)\end{matrix}\right.\)

Xét (*) ta có VT \(\ge3\) (1)

mà \(VP=\dfrac{1}{\sqrt{x+13}+4}\le\dfrac{1}{4}\) (2)

Từ (1) và (2) dễ thấy (*) vô nghiệm 

Hay x = 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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Ta có  A > 0 

Từ đó  \(A^2=2+\sqrt{2+\sqrt{2+...}}\Leftrightarrow A^2=2+A\Leftrightarrow A^2-A-2=0\)

\(\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\Leftrightarrow\orbr{\begin{cases}A+1=0\\A-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}A=-1\\A=2\end{cases}}\)

Do A > 0 nên A= 2

b, tương tự

c,\(C>2\)

Xét \(C^2=5+\sqrt{13+\sqrt{5+\sqrt{13...}}}\)

\(\left(C^2-5\right)^2=13+C\Leftrightarrow C^4-10C^2-C+12=0\Leftrightarrow\left(C^4-9C^2\right)-\left(C^2-9\right)-\left(C-3\right)=0\)

\(\Leftrightarrow\left(C-3\right)\left[\left(C+3\right)\left(C-1\right)\left(C+1\right)-1\right]=0\)

VÌ C> 2 =>  C-3 = 0 => C=3

\(=\sqrt[3]{25-52}=-3\)

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tham khảo! bài này mk làm ở đó hơi thieuus bạn chỉ cần + ... là đc