Помогите пж 1,2,3 номерУМОЛЯЮ!!!!

III)(1/b(abc+a+c)) - (1/(a+(1/(b+1/c))))(1/(a+(1/b)))=(1/(abbbc+ab+bc)) - (1/(a+(1/(b+ 1/c ))))((a+(1/b))/1)=(1/(bbac+ab+bc)) - ((a+1/b)/(a+(1/(b+1/c)))=(1/(bbac+ab+bc)) - ((a+ 1/b)/(a(b+1/c)/(b+1/c)+(1/b+1/c))=(1/(bbac+ab+bc)) - ((a+1/b)/((ab+a/c+1)/(b+1/c))=(1/ (bbac+ab+bc)) - (((a+1/b)/1)((b+1/c)/(ab+a/c+1))=(1/(bbac+ab+bc)) - (((a+1/b)(b+ 1/c))/(abc/c+a/c+c/c))=(1/(bbac+ab+bc)) - ((ab+b/b+a/c+1/bc)/((abc+a+c)/c))=(1/(b bac+ab+bc)) - (((ab+b/b+a/c+1/bc)/1)(c/(abc+a+c)))=(1/(bbac+ab+bc)) - (((ab+ b/b +a/c+1/bc)c)/(abc+a+c))=(1/(bbac+ab+bc)) - ((abc+cb/b+ac/c+c/bc)/(abc+a+c))=(1/b(abc+a+c)) - ((abc+cb/b+ac/c+c/bc)/(abc+a+c))=((1-abc-cb/b-ac/c-c/bc) /(abc +a +c))=(bc/bc-abc/bc-cb/bc-acb/bc-c/bc)/(abc+a+c))=((bc-abc-cb-acb-c)/bc)/(abc+a+c)=(bc(1-abc-c-a-1)/bc)/(abc+a+c)=(abc-c-a)/(abc+a+c)

II)(4a-((4a-1)/a))/(2-1/a)=(4a-4a+1)/a/(2a-1)/a=(4a- 4a+1)/aa/(2a-1)=(4a- 4a+1)/(2a-1) =((4a- 4a)-1)/(2a-1)
1)(4a-((4a-1)/a))=(4aa/a) - ((4a-1)/a)=4a/a - ((4a-1)/a)=(4a- 4a+1)/a
2)2 - 1/a=2a/a-1/a=(2a-1)/a
Найдём област возможных значений ((4a- 4a)-1)/(2a-1):
Решение: 2a-1