1,2(b+1,2)+(0,5-b)(b+0,5)-(b+1,3)(1,3-b)==1,2b+1,44+(0,5-b)(0,5+b)-(1,3+b)(1,3-b)==1,2b+1,44+0,5b^2-b^2-(1,3^2-b^2)==1,2b+1,44+0,5^2-b^2-((13/10)^2-b^2)==1,2b+1,44+0,5^2-b^2-(13/10)^2+b^2==1,2b+36/25+(1/2)^2-169/100==1,2b+36/25+1/4-169/100==1,2b+0=1,2bответ при b=5/6
1,2b=1,2*(-5/6)=-1
1,2(b+1,2)+(0,5-b)(b+0,5)-(b+1,3)(1,3-b)=
=1,2b+1,44+(0,5-b)(0,5+b)-(1,3+b)(1,3-b)=
=1,2b+1,44+0,5b^2-b^2-(1,3^2-b^2)=
=1,2b+1,44+0,5^2-b^2-((13/10)^2-b^2)=
=1,2b+1,44+0,5^2-b^2-(13/10)^2+b^2=
=1,2b+36/25+(1/2)^2-169/100=
=1,2b+36/25+1/4-169/100=
=1,2b+0=1,2b
ответ при b=5/6
1,2b=1,2*(-5/6)=-1