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Randomfack
@Randomfack
August 2021
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Срочно!!!!!!
Помогите, пожалуйста...
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Regent1828
Verified answer
[(a²+ab+bc-c²) / (b²-a²+2ac-c²)] : [(a²+ac) / (b²-ab+bc)] =
=[(a²-c²+ab+bc) / (b²-(a-c)
²
)] * [(b(b-a+c)) / (a(a+c))]=
=[((a-c)(a+c)+b(a+c))*b(b-a+c)] / ((b-a+c)(b+a-c)*a(a+c))=
=b(a+c)(b-a+c)(b+a-c) / a(a+c)(b-a+c)(b+a-c)=b/a
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Answers & Comments
Verified answer
[(a²+ab+bc-c²) / (b²-a²+2ac-c²)] : [(a²+ac) / (b²-ab+bc)] ==[(a²-c²+ab+bc) / (b²-(a-c)²)] * [(b(b-a+c)) / (a(a+c))]=
=[((a-c)(a+c)+b(a+c))*b(b-a+c)] / ((b-a+c)(b+a-c)*a(a+c))=
=b(a+c)(b-a+c)(b+a-c) / a(a+c)(b-a+c)(b+a-c)=b/a