Zadanie 1
a)
x^{-2} *(2x^{-1} +x^2)=\frac{1}{x^2}*(\frac{2}{x}+x^2)=\frac{2}{x^3}+\frac{x^2}{x^2}=\frac{2}{x^3}+1=\frac{2+x^3}{x^3}
zał. x\ne0
b)
(x^{-2} -x^{-1})(2x+x^2)=(\frac{1}{x^2}-\frac{1}{x})(2x+x^2)=\frac{2x}{x^2}+\frac{x^2}{x^2}-\frac{2x}{x}-\frac{x^2}{x}=
=\frac{2}{x}+1-2-x=\frac{2}{x}-1-x=\frac{2-x-x^2}{x}=\frac{-x^2-x+2}{x}
zał. x\ne0
c)
(y^{-3}+y^{-2} +y^{-1})(y^2 -y^3)=(\frac{1}{y^3}+\frac{1}{y^2}+\frac{1}{y})(y^2-y^3)=
=\frac{y^2}{y^3}+\frac{y^2}{y^2}+\frac{y^2}{y}-\frac{y^3}{y^3}+\frac{y^3}{y^2}+\frac{y^3}{y}=\frac{1}{y}+1+y-1+y+y^2=
=\frac{1}{y}+y^2=\frac{1+y^3}{y}
zał. y\ne0
d)
(y^2 +y^{-2})^2=(y^2+\frac{1}{y})^2=(y^2)^2+2*y^2*\frac{1}{y}+\frac{1}{y^2}=y^4+2y+\frac{1}{y^2}=\frac{y^6+2y^3+1}{y^2}
zał. y\ne0
e)
(3y^{-1} -y)^2=(\frac{3}{y}-y)^2=\frac{9}{y^2}-2*\frac{3}{y}*y+y^2=\frac{9}{y^2}-6+y^2=\frac{9-6y^2+y^4}{y^2}=
=\frac{y^4-6y^2+9}{y^2}
zał. y\ne0
f)
(x^{-1}+x)(x^{-1}-x)(x^{-2}+x^2)=[(x^{-1})^2-x^2] (x^{-2}+x^2)=(x^{-2}-x^2)(x^{-2}+x^2)=
=(x^{-2})^2-(x^2)^2=x^{-4}-x^4=\frac{1}{x^4}-x^4=\frac{1-x^8}{x^4}
zał. x\ne0