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[tex](x - y) {}^{2022} + (x - y) {}^{2021} = 0[/tex]
trebuie sa demonstrez că e 0​


Răspuns :

Ai în imagine rezolvarea exercițiului 3.

Vezi imaginea Аноним

[tex]\it x=\Big(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\Big)\cdot\dfrac{3}{2}=\dfrac{1}{2}\cdot\dfrac{3}{2}+\dfrac{1}{3}\cdot\dfrac{3}{2}-\dfrac{1}{6}\cdot\dfrac{3}{2}=\dfrac{3}{4}+\dfrac{^{2)}1}{2}-\dfrac{1}{4}=\dfrac{3+2-1}{4}=1\\ \\ \\ y=16^2:(2^2)^3:2=(2^4)^2:2^6:2=2^8:2^6:2=2^{8-6-1}=2[/tex]

[tex]\it (x-y)^{2022}+(x-y)2021=(x-y)^{2021}(x-y+1)=(1-2)^{2021}(1-2+1)=\\ \\ =(-1)^{2021}\cdot0=0[/tex]