A²-5a/3a²-48 - 3a-10/3a²-48 + 6/3a²-48= (a-4b)²/ab - вопрос №60062365348310348634844325 от 788SOS111 26.10.2020 21:37

Question

Алгебра: A²-5a/3a²-48 - 3a-10/3a²-48 + 6/3a²-48= (a-4b)²/ab - (a+4b)²/ab= 3a-2ab/ab-5b² + 15b-10b²/5b²-ab=

788SOS111 · Answer

A²-5a/3a²-48 - 3a-10/3a²-48 + 6/3a²-48=(a²-5a-3a+10+6)/(3a²-48)=
=(a²-8a+16)/(3a²-48)=(a-4)²/[3(a-4)(a+4)]=(a-4)/(3a+12)

(a-4b)²/ab - (a+4b)²/ab=(a-4b-a-4b)(a-4b+a+4b)/ab=-8b*2a/ab=-16

3a-2ab/ab-5b² + 15b-10b²/5b²-ab= (3a-2ab)/b(a-5b)-(15b-10b²)/b(a-5b)=
(3a-2ab-15b+10b²)/b(a-5b)=[a(3-2b)-5b(3-2b)]/b(a-5b)=
=(a-5b)(3-2b)/b(a-5b)=(3-2b)/b