\(D=\sqrt{\frac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\frac{\left(5-2\sqrt{6}\right)^2}{25-24}}=5+2\sqrt{6}+5-2\sqrt{6}=10\)

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

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

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

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

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

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

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

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

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

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

\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)

\(\Rightarrow A=\sqrt{12}\)

b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)

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

a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)

\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)

\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)

c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)

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

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

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

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

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

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

\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}=5\)

\(\sqrt{14+6\sqrt{5}}-\sqrt{\frac{\sqrt{5-2}}{\sqrt{5}+2}}=5\)

\(=\sqrt{9+2.3.\sqrt{5}+5}-\sqrt{\frac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}=\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{\frac{\left(\sqrt{5}-2\right)^2}{5-4}}\)

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

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

Đặt \(B=\frac{\sqrt{11+\sqrt{5}}+\sqrt{11-\sqrt{5}}}{\sqrt{11+2\sqrt{29}}}\)Ta có B>0

\(B^2=2\Rightarrow B=\sqrt{2}\)

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

Đặt \(x=\sqrt{\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}}>0\)

\(x^2=\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}+2\sqrt{\dfrac{25-24}{25-6}}=\dfrac{74}{19}+\dfrac{2\sqrt{19}}{19}\)

\(\Rightarrow x^2=\dfrac{74+2\sqrt{19}}{19}\Rightarrow x=\sqrt{\dfrac{74+2\sqrt{19}}{19}}\)

Ko thể rút gọn thêm nữa (có thể trục căn thức ở mẫu)