\(a,\\ \Leftrightarrow x^2+3x-x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ b,\\ \Leftrightarrow\left[{}\begin{matrix}4x=-4\\y=x+5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=-1+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
\(c,=\dfrac{4}{3-\sqrt{5}}-\dfrac{2\sqrt{5}}{2}-5\\ =\dfrac{8-2\sqrt{5}-5.2\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\\ =\dfrac{-12+4\sqrt{5}}{3-\sqrt{5}}=\dfrac{-4\left(3-\sqrt{5}\right)}{3-\sqrt{5}}=-4\\ d,\\ \Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+1\right)^2}-\dfrac{1}{x+1}-3=0\\ \Leftrightarrow\left(x+2\right)^2-\left(x+1\right)-3\left(x+1\right)^2=0\\ \Leftrightarrow x^2-3x^2+4x-x-6x+4-1-3=0\\ \Leftrightarrow-2x^2-3x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\left(tm\right)\)
a. \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\) ( vi-et )
b.\(\Leftrightarrow\left\{{}\begin{matrix}4x=-4\\x-y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\-1-y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\)
d.\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+1\right)^2}-\dfrac{1}{x+1}-3=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x+1\right)-3\left(x+1\right)^2}{\left(x+1\right)^2}=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)-3\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2+4x+4-x-1-3x^2-6x-3=0\)
\(\Leftrightarrow-2x^2-3x=0\)
\(\Leftrightarrow x\left(-2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)