a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

\(=2\sqrt{2}\)

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$

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

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

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

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

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

\(a,P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{4-x}\right):\dfrac{x+5\sqrt{x}+6}{x-4}\left(dk:x\ge0,x\ne4\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{4x}{x-4}\right).\dfrac{x-4}{x+2\sqrt{x}+3\sqrt{x}+6}\)

\(=\dfrac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2+4x}{x-4}.\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4x+8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}+3}\)

\(b,x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{4}}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\\ =\sqrt{5}+2-\sqrt{5}+2\\ =4\)

Khi \(x=4\Rightarrow P=\dfrac{4\sqrt{4}}{\sqrt{4}+3}=\dfrac{4.2}{2+3}=\dfrac{8}{5}\)

\(c,P=2\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+3}=2\Leftrightarrow\dfrac{4\sqrt{x}-2\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=0\Leftrightarrow2\sqrt{x}-6=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)

\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\)

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

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

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

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

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

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

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

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

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

\(=\sqrt{43+30\sqrt{2}}\)

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

\(\sqrt{4-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\) 

\(=\sqrt{2+2\sqrt{2}.\sqrt{1}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2\sqrt{8}.\sqrt{1}+1}\) 

\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{8}-1\right)^2}\)

\(=\sqrt{2}-1+2-\sqrt{2}+\sqrt{8}-1\) 

\(=\sqrt{8}=2\sqrt{2}\)

P/s Nhi Cái Hằng Đẳng Thức cuối phân tích ra căn 8 là j sao ko ra thẳng 2 căn 2 luôn????

\(\sqrt{4-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}\) 

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

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

P/s mình có cách phân tích ra hằng đẳng thức easy hơn bạn Nhi nè :)))

\(2ab=2\sqrt{2}\Rightarrow ab=\sqrt{2}\Rightarrow\left(\sqrt{2}-1\right)^2\) Đó bạn chỉ cần lấy 2ab với cái 2ab ở trong hằng đẳng thức chia cả 2 vế cho 2 là đã có thể biết được hằng đẳng thức nhé

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

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

c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)

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

\(\sqrt{10+2\sqrt{17-4\sqrt{9+4\sqrt{5}}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left|2+\sqrt{5}\right|}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left(2+\sqrt{5}\right)}}\)

\(=\sqrt{10+2\sqrt{17-8-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{10+2\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)

\(=\sqrt{10+2\sqrt{\left(2-\sqrt{5}\right)^2}}\)

\(=\sqrt{10+2\cdot\left|2-\sqrt{5}\right|}\)

\(=\sqrt{10+2\cdot\left(-2+\sqrt{5}\right)}\)

\(=\sqrt{10+-4+2\sqrt{5}}\)

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

\(=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot1+1^2}\)

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

\(=\left|\sqrt{5}+1\right|\)

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

\(=\sqrt{10+2\sqrt{17-4\sqrt{5+4\sqrt{5}+4}}}\)

\(=\sqrt{10+2\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\)

\(=\sqrt{10+2\sqrt{17-4\cdot\left|\sqrt{5}+2\right|}}\)

\(=\sqrt{10+2\sqrt{17-4\left(\sqrt{5}+2\right)}}\) (vì \(\sqrt{5}+2>0\))

\(=\sqrt{10+2\sqrt{17-4\sqrt{5}-8}}\)

\(=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\\ =\sqrt{10+2\sqrt{5-4\sqrt{5}+4}}\\ =\sqrt{10+2\sqrt{\left(\sqrt{5}-2\right)^2}}\\ =\sqrt{10+2\cdot\left|\sqrt{5}-2\right|}\)

\(=\sqrt{10+2\cdot\left(\sqrt{5}-2\right)}\) (vì \(\sqrt{5}-2>0\))

\(=\sqrt{10+2\sqrt{5}-4}\\ =\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}\\ =\left|\sqrt{5}+1\right|\)

\(=\sqrt{5}+1\) (vì \(\sqrt{5}+1>0\))

a.

\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)

c.

\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)

d.

\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)

Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì

\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)

Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)

Vậy \(\left(a;b\right)=\left(9;4\right)\)