Câu 1
\(\left[ \begin{array}{l}4x+7y=18\\3x-9y=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}12x+21y=54\\12x-36y=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}57y=50\\3x-9y=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}y=\frac{50}{57}\\3x-9.\frac{50}{57}=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}y=\frac{50}{57}\\3x=\frac{169}{19}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}y=\frac{50}{57}\\y=\frac{169}{57}\end{array} \right.\)
Câu 2
$\frac{2}{\sqrt[]{5}-1}=\frac{2(\sqrt[]{5}+1)}{(\sqrt[]{5}-1)(\sqrt[]{5}+1)}=\frac{2(\sqrt[]{5}+1)}{(\sqrt[]{5})^2-1^2}=\frac{2(\sqrt[]{5}+1)}{5-1}=\frac{2(\sqrt[]{5}+1)}{4}=\frac{\sqrt[]{5}+1}{2}$
Câu 3
\(\left[ \begin{array}{l}x-y=1\\2x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x-y=1\\2x=-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x-y=1\\x=\frac{-3}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\frac{-3}{2}-y=1\\x=\frac{-3}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}y=\frac{-5}{2}\\x=\frac{-3}{2}\end{array} \right.\)
