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a)[4x²-1]/[2x³+x²}
D:2x³+x²≠0 x²(2x+1)≠0 x≠0 i x≠-1/2
D;x∈R-{-1/2,0}
[4x²-1]/[2x³+x²}=(2x-1)(2x+1)/x²(2x+1)=[2x-1]/x²
b({x³+4x]/[x^5-16x] D:x^5-16x≠0
x(x^4-16)≠0
x≠0 albo x^4-16≠0 (x²-4)(x²+4)≠0
x≠-2 x≠2
D:x∈R-{-2,0,2}
[x³+4x]/[x^5-16x]=x(x²+4)/x(x^4-16)=x(x²+4)/x(x²-4)(x²+4)=1/[x²-4]
c) [x²-2x+1]/[x³-1] D:x³≠1 D:x∈R-{1}
{x²-2x+1}/x³-1]=(x-1)(x-1)/(x-1)(x²+x+1) =(x-1)/[x²+x+1)
d) [x²+2x-8}/[x²-4x+4] D:x²-4x+4≠0 (x-2)²≠0 x≠2 D:x∈R-{2}
{x²+2x-8)/[x²-4x+4] z licznika obliczymyΔ=4+32=36
√Δ=6
x1=-2-6/2 x2=-2+6/2
x1=-4 x2=2
stosujemy postać iloczynow ą trójmianu kwadratowego
(x+4)(x-2)/(x-2)²=[x+4]/[x-2]