1)
\(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{1+\sqrt{5}}{4}-\dfrac{4\sqrt{5}-4}{4}\) \(=\dfrac{5-3\sqrt{5}}{4}\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\9x+3y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7\\y=4-3x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(1;1\right)\)
1: Rút gọn
Ta có: \(\dfrac{1}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}\)
\(=\dfrac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-4\sqrt{5}+4}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{5-3\sqrt{5}}{4}\)
2)
Ta có: \(\left\{{}\begin{matrix}2x+3y=5\\3x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+9y=15\\6x+2y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7y=7\\3x+y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\3x=4-y=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)