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kutsenckosofia
@kutsenckosofia
July 2022
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sedinalana
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1)1-(b²+c²-a²)/2bc=(2bc-b²-c²+a²)/2bc=(a²-(b-c)²)/2bc
2)1/a-1/(b-c)=(b-c-a)/[a(b-c)]
3)1/a+1/(b-c)=(b-c+a)/[a(b-c)]
4)(b-c-a)/[a(b-c)] :(b-c+a)/[a(b-c)]=(b+c-a)/[a(b-c)]*a(b-c)/(b-c+a)=
=(b-c-a)/(b-c+a)
5)(b-c-a)/(b-c+a) *(a-b+c)(a+b-c)/2bc*abc/(b-a-c)=a(a-b+c)/2=0,5a(a-b+c)
a=1,2;b=0,5;c=1,3
1,2*(1,2-0,5+1,3)/2=0,6*2=1,2
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Answers & Comments
Verified answer
1)1-(b²+c²-a²)/2bc=(2bc-b²-c²+a²)/2bc=(a²-(b-c)²)/2bc2)1/a-1/(b-c)=(b-c-a)/[a(b-c)]
3)1/a+1/(b-c)=(b-c+a)/[a(b-c)]
4)(b-c-a)/[a(b-c)] :(b-c+a)/[a(b-c)]=(b+c-a)/[a(b-c)]*a(b-c)/(b-c+a)=
=(b-c-a)/(b-c+a)
5)(b-c-a)/(b-c+a) *(a-b+c)(a+b-c)/2bc*abc/(b-a-c)=a(a-b+c)/2=0,5a(a-b+c)
a=1,2;b=0,5;c=1,3
1,2*(1,2-0,5+1,3)/2=0,6*2=1,2