Zadanie 5
a)
\frac{x-1}{3x^2}*\frac{5x^3}{x^2-1}=\frac{x-1}{3}*\frac{5x}{(x-1)(x+1)}=\frac{5x}{3(x+1)}=\frac{5x}{3x+3}
3x^2\ne0 => x\ne0
x^2-1\ne0 => x^2\ne 1
x\ne-1 i x\ne 1
b)
\frac{x^2-16}{x^2-x-6}:\frac{x^2-x-20}{x^2-x-6}=\frac{x^2-16}{x^2-x-6}*\frac{x^2-x-6}{x^2-x-20}=\frac{x^2-16}{x^2-x-20}=\frac{(x-4)(x+4)}{(x+4)(x-5)}=\frac{x-4}{x-5}
x^2-x-6\ne0
\Delta=(-1)^2-4*1*(-6)=1+24=25
\sqrt\Delta=5
x_1=\frac{1-5}{2}=-2
x_2=\frac{1+5}{2}=3
x\ne -2 i x\ne 3
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x^2-x-20\ne 0
\Delta=1-4*1*(-20)=81
\sqrt\Delta=9
x_1=\frac{1-9}{2}=-4
x_2=\frac{1+9}{2}=5
x \in R - {-4, -2, 3, 5}