Question

ДАЮ 100 БАЛЛОВ
упростите выражения c - d/c ^2 +cd - c/d^2 +cd ÷ d^2 /c^ 3- cd^2 + 1/ c +d​​

Answer 1

Ответ:

$1)\ \ \dfrac{p-q}{p}\cdot \Big(\dfrac{p}{p-q}+\dfrac{p}{q}\Big)=\dfrac{p-q}{p}\cdot \dfrac{pq+p^2-pq}{q(p-q)}=\dfrac{p^2}{pq}=\dfrac{p}{q}$

$2)\ \ \Big(\dfrac{2a-b}{4a^2+2ab}-\dfrac{2a}{b^2+2ab}\Big):\Big(\dfrac{b^2}{8a^3-2ab^2}+\dfrac{1}{2a+b}\Big)=\\\\\\=\Big(\dfrac{2a-b}{2a(2a+b)}-\dfrac{2a}{b(b+2a)}\Big):\Big(\dfrac{b^2}{2a(2a-b)(2a+b)}+\dfrac{1}{2a+b}\Big)=\\\\\\=\dfrac{2ab-b^2-4a^2}{2ab(2a+b)}:\dfrac{b^2+4a^2-2ab}{2a(2a-b)(2a+b)}=\\\\\\=\dfrac{-(b^2+4a^2-2ab)}{2ab(2a+b)}\cdot \dfrac{2a(2a-b)(2a+b)}{b^2+4a^2-2ab}=-\dfrac{2a-b}{b}=\dfrac{b-2a}{b}$

$P.S.\ \ \Big(\dfrac{c-d}{c^2+cd}-\dfrac{c}{d^2+cd}\Big):\Big(\dfrac{d^2}{c^3-cd^2}+\dfrac{1}{c+d}\Big)=\\\\\\=\Big(\dfrac{c-d}{c(c+d)}-\dfrac{c}{d(c+d)}\Big):\Big(\dfrac{d^2}{c(c-d)(c+d)}+\dfrac{1}{c+d}\Big)=\\\\\\=\dfrac{cd-d^2-c^2}{c\, d(c+d)}:\dfrac{d^2+c^2-cd}{c(c-d)(c+d)}=\\\\\\=\dfrac{cd-d^2-c^2}{c\, d(c+d)}\cdot \dfrac{c(c-d)(c+d)}{-(cd-d^2-c^2)}=-\dfrac{c-d}{d}=\dfrac{d-c}{d}$