[tex](x-y)(x+y)-(a-x+y)(a-x-y)-a(2x-a)=x^2-y^2-((a-x)^2-y^2)-2ax-a^2=x^2-y^2-(a^2-2ax+x^2-y^2)-2ax-a^2=x^2-y^2-a^2+2ax-x^2+y^2-2ax-a^2=0[/tex]
Отже, дійсно, [tex](x-y)(x+y)-(a-x+y)(a-x-y)-a(2x-a)=0[/tex]
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[tex](x-y)(x+y)-(a-x+y)(a-x-y)-a(2x-a)=x^2-y^2-((a-x)^2-y^2)-2ax-a^2=x^2-y^2-(a^2-2ax+x^2-y^2)-2ax-a^2=x^2-y^2-a^2+2ax-x^2+y^2-2ax-a^2=0[/tex]
Отже, дійсно, [tex](x-y)(x+y)-(a-x+y)(a-x-y)-a(2x-a)=0[/tex]