\(a.A=\left(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\dfrac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}-1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{0}{1-3}\right):\dfrac{1}{\sqrt{3}}\) \(=0:\dfrac{1}{\sqrt{3}}=0\)
b. B được xác định ⇔ x > 0 ; \(x\ne1\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\)
\(B=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\).
\(B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c. Giả Sử A = \(\dfrac{1}{6}B\)
⇔ 0 = \(\dfrac{1}{6}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
⇔ 0=\(\dfrac{\sqrt{x}-1}{6\sqrt{x}}\)
⇔0 = \(\sqrt{x}-1\)
⇔x = 1(không thỏa mãn)
⇒ A ≠ \(\dfrac{1}{6}B\)
Vậy A ≠ \(\dfrac{1}{6}B\) (Do x không có giá trị nào thỏa mãn)
