[tex]zad.43\\zal. y\neq z,~~z\neq x,~~x\neq y\\\\zadanie~~polega~~na~~wykazaniu~~L=P\\\\P=0\\\\L=\frac{x+y}{(y-z)(z-x)} -\frac{y+z}{(x-z)(x-y)} -\frac{z+x}{(y-x)(y-z)} \\\\L=\frac{x+y}{(y-z)(z-x)} -\frac{y+z}{(-1)(z-x)(-1)(y-x)} -\frac{z+x}{(y-x)(y-z)}\\\\L=\frac{x+y}{(y-z)(z-x)} -\frac{y+z}{(z-x)(y-x)} -\frac{z+x}{(y-x)(y-z)}\\\\L=\frac{(x+y)(y-x)-(y+z)(y-z)-(z+x)(z-x)}{(y-z)(z-x)(y-x)}\\\\L= \frac{(y+x)(y-x)-(y+z)(y-z)-(z+x)(z-x)}{(y-z)(z-x)(y-x)}\\\\[/tex]
[tex]L=\frac{y^{2}-x^{2}-(y^{2}-z^{2})-(z^{2}-x^{2}) }{(y-z)(z-x)(y-x)}\\\\L=\frac{y^{2}-x^{2}-y^{2}+z^{2}-z^{2}+x^{2}}{(y-z)(z-x)(y-x)}\\\\L=\frac{0}{(y-z)(z-x)(y-x)}\\\\L=0\\\\L=0~~\wedge~~P=0~~\rightarrow ~~L=P~~ cbdu.[/tex]
Korzystałam ze wzoru skróconego mnożenia:
[tex](a+b)(a-b)=a^{2} -b^{2}[/tex]
