Bài 1

a) Đặt VT = A

<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

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

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

<=> 2A = \(\left(5-3\right)^2=4\)

<=> A = 2

b) Đặt VT = B

<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)

<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)

<=> B = 8 

Bài 2

Đặt VT = A

<=> A² = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)

<=> A² = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)

<=> \(A=\sqrt{\sqrt{5}+1}\)

\(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}-\dfrac{3}{3-\sqrt{6}}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)

\(=\left(\dfrac{2+2\sqrt{6}}{5}+\dfrac{6+3\sqrt{6}}{2}-3-\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)

\(=\dfrac{4+4\sqrt{6}+30+15\sqrt{6}-30-10\sqrt{6}}{10}\cdot\dfrac{5}{9\sqrt{6}+4}\)

\(=\dfrac{1}{2}\)

bình phương 2 vế dc pt tương đương

\(-\left(4x^2-15x+8\right)\left(4x^2-11x+3\right)=0\)

A= căn (5-2 (căn 5) +1)-căn (5+2 (căn 5) +1)

=căn ((căn 5)-1)^2 -căn ((căn 5)+1)^2

=l (căn 5) -1l  -   l (căn 5) +1l

=căn 5 -1 -căn 5 -1 

=-2

A,  biến đổi 6= căn bậc hai của 5 + 1 -> hằng đẳng thức

Tính tiếp sẽ ra

\(x^2=6+2\sqrt{2}+2\sqrt{\left[\left(3+\sqrt{2}\right)+\left(\sqrt{3}+\sqrt{6}\right)\right].\left[\left(3+\sqrt{2}\right)-\left(\sqrt{3}+\sqrt{6}\right)\right]}\)

\(=6+2\sqrt{2}+2\sqrt{11+6\sqrt{2}-\left(9+6\sqrt{2}\right)}=6+2\sqrt{2}+2\sqrt{2}=6+4\sqrt{2}=\left(\sqrt{2}+2\right)^2\)

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

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