Question

Соч по алгебре 7 класс 4 четверть

Answer 1

$1)\; \; \dfrac{14x^4y}{49x^3y^2}=\dfrac{2x}{7y}\\\\\\2)\; \; \dfrac{x-y}{7}-\dfrac{x}{7}=\dfrac{x-y-x}{7}=-\dfrac{y}{7}$

$3)\; \; \dfrac{3a-9}{a+2}:\dfrac{a^2-9}{a^2-4}=\dfrac{3(a-3)}{(a+2)}\cdot \dfrac{(a-2)(a+2)}{(a-3)(a+3)}=\dfrac{3\, (a-2)}{a+3}$

$4)\; \; \dfrac{2a}{a-5}-\dfrac{5}{a+5}+\dfrac{2a^2}{25-a^2}=\dfrac{2a}{a-5}-\dfrac{5}{a+5}+\dfrac{2a^2}{(5-a)(5+a)}=\\\\\\=\dfrac{2a(a+5)-5(a-5)-2a^2}{(a-5)(a+5)}=\dfrac{2a^2+10a-5a+25-2a^2}{(a-5)(a+5)}=\\\\\\=\dfrac{5a+25}{(a-5)(a+5)}=\dfrac{5(a+5)}{(a-5)(a+5)}=\dfrac{5}{a-5}$

$5)\; \; a=-0,4\; \; ,\; \; \dfrac{4a^2+8}{a^3+1}-\dfrac{4}{a^2-a+1}-\dfrac{1}{a+1}=\\\\\\=\dfrac{4a^2+8}{(a+1)(a^2-a+1)}-\dfrac{4}{a^2-a+1}-\dfrac{1}{a+1}=\\\\\\=\dfrac{4a^2+8-4(a+1)-a^2+a-1}{(a+1)(a^2-a+1)}=\dfrac{3a^2-3a+3}{(a+1)(a^2-a+1)}=\dfrac{3(a^2-a+1)}{(a+1)(a^2-a+1)}=\\\\\\=\dfrac{3}{a+1}=\dfrac{3}{-0,4+1}=\dfrac{3}{0,6}=5$