a)
\(A=\left(\frac{1}{1-\sqrt{3}}-\frac{1}{1+\sqrt{3}}\right):\frac{1}{\sqrt{3}}\\ =\left(\frac{1+\sqrt{3}-1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\frac{1}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{1-\left(\sqrt{3}\right)^2}:\frac{1}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{-2}\cdot\sqrt{3}=-3\)
\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}\left(ĐK:x>0;x\ne1\right)\\ =\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x-1}\right)}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b)
\(A=\frac{1}{6}B\Leftrightarrow-3=\frac{1}{6}\cdot\frac{\sqrt{x}-1}{\sqrt{x}}\\ \Leftrightarrow-18=\frac{\sqrt{x}-1}{\sqrt{x}}\Leftrightarrow-18\sqrt{x}=\sqrt{x}-1\\ \Leftrightarrow-19\sqrt{x}=-1\\ \Leftrightarrow\sqrt{x}=\frac{1}{19}\Rightarrow x=\frac{1}{361}\)
Vậy với x = \(\frac{1}{361}\)thì \(A=\frac{1}{6}B\)
Có gì sai mọi người góp ý nha!
