Question

As dori si eu rezolvarea acestor 2 probleme. Daca se poate sa imi si explicati cum ati rezolvat. Mersi!

Answer 1

$Inecuatii~de~gradul~2:\\ \\ 1)ax^2+(a-1)x+a-1<0\\ \\ Conditii:~a<0~si~\Delta<0.\\ \\ \Delta=(a-1)^2-4(a-1)=a^2-2a+1-4a+4=a^2-6a+5\\ \\ Deci,~a^2-6a+5<0\\ \\ Asociezi~ecuatia~de~gradul~2~cu~radacinile~a_1=5~si~a_2=1.\\ \\ Faci~tabelul~si~obtii~a\in~(1,5).\\ \\ Dar,~a<0,~deci~\boxed{a\in\varnothing.} \\ \\ 2)ax^2+x-a>0\\ \\ Conditii:~a>0~si~\Delta<0\\ \\ \Delta=4a^2+1\geq 0~dar~\Delta<0~\Rightarrow~\Delta=0~\Leftrightarrow~a^2=\frac{1}{4} ~\Leftrightarrow~a=\{\frac{1}{4},-\frac{1}{4} \}$

Answer 2

Poza conține rezolvarea.