Đáp án:
a. $S={3}$
b.$S={2;-3}$
c.$(x;y)=(2;-3)$
d. $P=5$
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ 3x-5=x+1\\ \Leftrightarrow \ 2x=6\\ \Leftrightarrow x=3\\ Vậy\ S=\{3\}\\ b.\ x^{2} +x-6=0\\ \Leftrightarrow ( x-2)( x+3) =0\\ \Leftrightarrow [_{x=-3}^{x=2}\\ Vậy\ S=\{2;-3\}\\ c.\ \{_{x+y=-1( 2)}^{x-2y=8\ ( 1)}\\ ( 1) -( 2) :\ -3y=9\\ \Leftrightarrow y=-3,\ thay\ ( 2)\\ x=-1-y=-1--3=2\\ Vậy\ ( x,y) =( 2;-3)\\ d.\ \frac{\sqrt{5}}{\sqrt{5} -2} -2\sqrt{5} =\frac{\sqrt{5}\left(\sqrt{5} +2\right)}{5-4} -2\sqrt{5}\\ =5+2\sqrt{5} -2\sqrt{5} =5\\ \end{array}$
