b1/ $\left(\dfrac{\sqrt{14}-\sqrt 7}{1-\sqrt 2}-\dfrac{\sqrt{15}-\sqrt 5}{1-\sqrt 3}\right):\dfrac{1}{\sqrt 7+\sqrt 5}\\\quad =\left(\dfrac{\sqrt{7.2}-\sqrt 7}{1-\sqrt 2}-\dfrac{\sqrt{5.3}-\sqrt 5}{1-\sqrt 3}\right).(\sqrt 7+\sqrt 5)\\\quad =\left(\dfrac{\sqrt 7.\sqrt 2-\sqrt 7}{1-\sqrt 2}+\dfrac{\sqrt 5-\sqrt 5.\sqrt 3}{1-\sqrt 3}\right).(\sqrt 7+\sqrt 5)\\\quad =\left(-\dfrac{\sqrt 7(1-\sqrt 2)}{1-\sqrt 2}+\dfrac{\sqrt 5(1-\sqrt 3)}{1-\sqrt 3}\right).(\sqrt 7+\sqrt 5)\\\quad =(-\sqrt 7+\sqrt 5)(\sqrt 7+\sqrt 5)\\\quad =(\sqrt 5-\sqrt 7)(\sqrt 5+\sqrt 7)\\\quad =(\sqrt 5)^2-(\sqrt 7)^2\\\quad =5-7\\\quad =-2$
Vậy $\left(\dfrac{\sqrt{14}-\sqrt 7}{1-\sqrt 2}-\dfrac{\sqrt{15}-\sqrt 5}{1-\sqrt 3}\right):\dfrac{1}{\sqrt 7+\sqrt 5}=-2$
a2/ ĐK: $x\in\Bbb R$
$A\,=\sqrt{3x^2-4x\sqrt 3+4}\\\quad =\sqrt{(\sqrt 3x)^2-2.\sqrt 3x.2+2^2}\\\quad =\sqrt{(\sqrt 3x-2)^2}\\\quad =|\sqrt 3x-2|$
Thay $x=\sqrt 3+\dfrac{1}{\sqrt 3}$ (thỏa mãn điều kiện) vào $A$
$A\,=\left|\sqrt 3.\left(\sqrt 3+\dfrac{1}{\sqrt 3}\right)-2\right|\\\quad =|3+1-2|\\\quad =|2|\\\quad =2$
Vậy $A=2$ tại $x=\sqrt 3+\dfrac{1}{\sqrt 3}$
