a) A xác định <=>\(\left\{{}\begin{matrix}x-1\ge0\\3-x\le0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\)
<=>\(1\le x\le3\)
b)\(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{\sqrt{5}+1}=\dfrac{3+\sqrt{5}}{9-5}-\dfrac{\sqrt{5}-1}{5-1}=\dfrac{3+\sqrt{5}-\sqrt{5}+1}{4}=\dfrac{4}{4}=1\)
a.ĐK:\(x\ge1;x\ge-3\)
b.\(\dfrac{\sqrt{5}+1-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}-2}{3\sqrt{5}+3-2\sqrt{5}-\sqrt{5}}\) ( bạn quy đồng )
\(\dfrac{2\sqrt{5}-2}{3}\)
a: ĐXKĐ: \(\left\{{}\begin{matrix}x-1>=0\\3-x>=0\end{matrix}\right.\Leftrightarrow1\le x\le3\)
b: \(=\dfrac{3+\sqrt{5}}{4}-\dfrac{\sqrt{5}-1}{4}=\dfrac{3+\sqrt{5}-\sqrt{5}+1}{4}=1\)
