Question

Помогите пжжжж, за четверть выходит 2​

Answer 1

$3a) \ \dfrac{a}{5a+1} \\\\5a+1\neq 0\\\\5a\neq -1\\\\a\neq -0,2\\\\Otvet:\boxed{a\in(-\infty \ ; \ -0,2) \ \cup \ (-0,2 \ ; \ +\infty)}\\\\\\3b) \ \dfrac{12+x}{8-8x+2x^{2} } \\\\8-8x+2x^{2} \neq 0\\\\x^{2} -4x+4\neq 0\\\\(x-2)^{2} \neq 0\\\\x-2\neq 0\\\\x\neq 2\\\\Otvet:\boxed{x\in(-\infty \ , \ 2) \ \cup \ (2 \ ; \ +\infty)}$

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

$2) \ \dfrac{4(a-2)(a+2)}{(a-4)(a-1)} \cdot\dfrac{a-4}{a-2} =\dfrac{4(a+2)}{a-1} \\\\\\3) \ \dfrac{4(a+2)}{a-1} +\dfrac{4a\cdot(a-1)}{a^{2} -2a+1} =\dfrac{4(a+2)}{a-1} +\dfrac{4a\cdot(a-1)}{(a-1)^{2} } =\\\\\\=\dfrac{4(a+2)}{a-1} +\dfrac{4a}{a-1} =\dfrac{4a+8+4a}{a-1} =\dfrac{8a+8}{a-1} =\dfrac{8(a+1)}{a-1} \\\\\\4) \ \dfrac{8(a+1)}{a-1} \cdot\dfrac{a-1}{a+1} =\boxed8$