1) x² *(4x²-16)=0 x1=x2=0 , x2=-2 , x4=2
2) x*(4x²+11x+6)=0
3)(x³-x-20x+20= (x³-x)-20(x-1)=x*(x²-1)-20*(x-1)=x*(x-1)(x+1)-20*(x-1)=(x-1)*[x*(x+1)-20)]=(x-1)*(x²+x-20)=0
4)x²(x³+1)=0 <=> x²(x+1)*(x²-x+1)=x²(x+1)*(x²-x+1) a doua paranteza
are dterminantul mai mic ca 0 deci e mmai mare decat 0 solutii x1=x2=0, x3=-1
5)4x²(x-3)-(x-3)=0 =>(x-3)*(4x²-1)=0 x-3=0 x=3
4x²-1=0 => x2./3=+/-√1/4= +/-1/2
