a) f(x)=[tex]( \sqrt{3}- \sqrt{2})x-1,avem,f(x)=0,deci,( \sqrt{3}- \sqrt{2})x-1=0,adica,( \sqrt{3}- [/tex][tex] \sqrt{2})x=1,sau,x= \frac{1}{ \sqrt{3}- \sqrt{2} }= \frac{ \sqrt{3}+ \sqrt{2} }{ (\sqrt{3}- \sqrt{2})( \sqrt{3}+ \sqrt{2}) }= \frac{ \sqrt{3}+ \sqrt{2} }{3-2}= \sqrt{3}+ \sqrt{2} [/tex]
b) [tex]( \sqrt{3} + \sqrt{2})f(x)+ \sqrt{3} - \sqrt{2}=0 [/tex], inlocuim pe f(x) si tinem cont din nou de formula (a-b)(a+b)=a²-b², vom avea:
[tex] (\sqrt{3} + \sqrt{2}) [ (\sqrt{3} - \sqrt{2})x-1]+ \sqrt{3} - \sqrt{2}=0,
[/tex], [tex](\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2})x-1(\sqrt{3} + \sqrt{2})+\sqrt{3} - \sqrt{2}=0[/tex],
Sau: (3-2)x-[tex] -\sqrt{3} - \sqrt{2} + \sqrt{3} - \sqrt{2}=0, [/tex], reducem termeni asemenea trecem in dreapta 2 radicalul din doi care a ramas si ⇒ x=2√2.
