Задание

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Упрости выражения.

\dfrac{1}{b(a+b)} \dfrac{16b}{a^2(a+b)} \dfrac{(a^2-8ab+16b^2)\cdot ab^2}{a^2b(a+b)\cdot (a^2-8ab+16b^2)} \dfrac{b}{a(a+b)}

1) \bigg( \dfrac{1}{ab+b^2}-\dfrac{8}{a^2+ab}+\dfrac{16b}{a^3+a^2b} \bigg) : \bigg( \dfrac{a}{b^2} - \dfrac{8}{b} + \dfrac{16}{a} \bigg) = \bigg(-\dfrac{8}{a(a+b)}+\bigg) :\dfrac{a^2-8ab+16b^2}{ab^2}==.

\dfrac{c(b+c)\cdot b \cdot (1+b^2-bc)}{(b+c)(b^2-bc)} \dfrac{cb(1+b^2-bc)}{b(b-c)} \dfrac{c+b^2c-bc^2-c}{b-c} bc

2) \bigg( \dfrac{bc+c^2}{b^2-bc}+bc+c^2 \bigg)\cdot \dfrac{b}{b+c} - \dfrac{c}{b-c}=(bc+c^2)\cdot \bigg( \dfrac{1}{b^2-bc}+1 \bigg) \cdot \dfrac{b}{b+c}-\dfrac{c}{b-c}=-\dfrac{c}{b-c}=-\dfrac{c}{b-c}==\dfrac{bc(b-c)}{b-c}=.

\dfrac{c+a+b}{abc} \dfrac{(c+a+b)\cdot ab}{abc \cdot (a^2-(b+c)^2)} \dfrac{1}{ac-bc-c^2}

3) \bigg( \dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac} \bigg) : \dfrac{a^2-b^2-c^2-2bc}{ab}=: \dfrac{a^2-(b^2+c^2+2bc)}{ab}==\dfrac{a+b+c}{c(a-b-c)(a+b+c)}=.

\bigg( \dfrac{a+a}{(a^3+a^2)+(a+1)} \bigg) \bigg( \dfrac{1}{a-1}-\dfrac{2a}{a^2(a-1)+(a-1)} \bigg) \dfrac{a+a+a+1}{(a+1)(a^2+1)} \dfrac{(a^2+2a+1)\cdot(a-1)(a^2+1)}{(a+1)(a^2+1)\cdot (a^2-2a+1)} \dfrac{a+1}{a-1}

4) \bigg( \dfrac{a^2+a}{a^3+a^2+a+1}+\dfrac{1}{a^2+1}\bigg):\bigg( \dfrac{1}{a-1}-\dfrac{2a}{a^3-a^2+a-1}\bigg)=:\bigg( \dfrac{1}{a-1}-\dfrac{2a}{(a^3-a^2)+(a-1)}\bigg)=\bigg(\cfrac{a^2+a}{a^2(a+1)+(a+1)}+\dfrac{1}{a^2+1}\bigg):=: \dfrac{a+1-2a}{(a-1)(a^2+1)}==\dfrac{(a+1)^2\cdot(a-1)}{(a-1)^2(a+1)}=.