Question

Помогите пожалуйста. Даю 50 баллов

Answer 1

$1) \ \dfrac{5a-4}{a-4} -\dfrac{a-5}{a-1} =\dfrac{(5a-4)(a-1)-(a-5)(a-4)}{(a-4)(a-1)} =\\\\\\=\dfrac{5a^{2}-5a-4a+4-a^{2} +4a+5a-20 }{(a-4)(a-1)} =\dfrac{4a^{2}-16 }{(a-4)(a-1)} =\\\\\\=\dfrac{4(a^{2}-4) }{(a-4)(a-1)} =\dfrac{4(a-2)(a+2) }{(a-4)(a-1)} \\\\\\2) \ \dfrac{4(a-2)(a+2) }{(a-4)(a-1)} \cdot\dfrac{a-4}{a-2} =\dfrac{4(a+2) }{a-1} \\\\\\3) \ \dfrac{4(a+2) }{a-1}+\dfrac{4a\cdot(a-1)}{a^{2}-2a+1 }\dfrac{4a+8}{a-1}+\dfrac{4a^{2} -4a}{(a-1)^{2}} =$

$=\dfrac{(4a+8)(a-1)+4a^{2}-4a }{(a-1)^{2} } =\dfrac{4a^{2} -4a+8a-8+4a^{2}-4a }{(a-1)^{2} } =\\\\\\=\dfrac{8a^{2} -8}{(a-1)^{2} } = \dfrac{8(a^{2} -1)}{(a-1)^{2} } = \dfrac{8(a-1)(a+1)}{(a-1)^{2} } = \dfrac{8(a+1)}{a-1} \\\\\\4) \ \dfrac{8(a+1)}{a-1} \cdot\dfrac{a-1}{a+1} =\boxed8$