Question

Алгебра, 8 класс
Пожалуйста помогите:(

Answer 1

а) $\frac{b-6}{4-b^{2} } +\frac{2}{2b-b^{2} } =\frac{b-6}{(2-b)(2+b)} +\frac{2}{b(2-b)} =\frac{(b-6)b}{(2-b)(2+b)b}} +\frac{2(2+b)}{b(2-b)(2+b)} =\frac{(b-6)b+2(2+b)}{b(2-b)(2+b)} =\frac{b^{2}-6b+4+2b}{b(2-b)(2+b)} =\frac{b^{2} -4b+4}{b(2-b)(2+b)} =\frac{(2-b)^{2} }{b(2-b)(2+b)} =\frac{2-b}{b(2+b)}$

б)$\frac{b}{ab-5a^{2} } -\frac{15b-25a}{b^{2} -25a^{2} } =\frac{b}{a(b-5a)} -\frac{5(3b-5a)}{(b-5a)(b+5a)} =\frac{b(b+5a)}{a(b-5a)(b+5a)}-\frac{5a(3b-5a)}{a(b-5a)(b+5a)}=\frac{b(b+5a)-5a(3b-5a)}{a(b-5a)(b+5a)}=\frac{b^{2} +5ab-15ab+25a^{2} }{a(b-5a)(b+5a)} =\frac{b^{2} -10ab+25a^{2} }{a(b-5a)(b+5a)}=\frac{(b-5a)^{2} }{a(b-5a)(b+5a)}=\frac{b-5a }{a(b+5a)}$

в$\frac{x-12a}{x^{2} -16a^{2} } -\frac{4a}{4ax-x^{2} } =\frac{x-12a}{(x-4a)(x+4a)} -\frac{4a}{x(4a-x)} =\frac{x(x-12a)}{x(x-4a)(x+4a)}-\frac{-4a(x+4a)}{x(x-4a)(x+4a)} =\frac{x(x-12a)-(-4a(x+4a))}{x(x-4a)(x+4a)} =\frac{x^{2} -12ax+4ax+16a^{2} }{x(x-4a)(x+4a)}=\frac{x^{2} -8ax+16a^{2} }{x(x-4a)(x+4a)}=\frac{(x-4a)^{2} }{x(x-4a)(x+4a)}=\frac{x-4a}{x(x+4a)}$

г)$\frac{a-30y}{a^{2}-100y^{2} } -\frac{10y}{10ay-a^{2} } =\frac{a-30y}{(a-10y)(a+10y) } -\frac{10y}{a(10y-a) }=\frac{a(a-30y)}{a(a+10y)(a-10y) } -\frac{-10y(a+10y)}{a(a+10y)(a-10y)}=\frac{a(a-30y)-(-10y(a+10y))}{a(a+10y)(a-10y) } =\frac{a^{2} -30ay+10ay+100y^{2} }{a(a+10y)(a-10y)} =\frac{a^{2} -20ay+100y^{2} }{a(a+10y)(a-10y)}=\frac{(a-10y)^{2} }{a(a+10y)(a-10y)}=\frac{a-10y}{a(a+10y)}=\frac{a-10y}{a^{2} +10ay}$