\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)

    \(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)

    \(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)

    \(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)

    \(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)

    \(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)

Đề lỗi rồi chứ mình ko rút gọn đc nữa

1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)

\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)

\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)

\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)

\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)

\(VT=\dfrac{2a^3-a^2b-a^2c-ab^2-ac^2+2b^3-b^2c-bc^2+2c^3}{(a+b)(b+c)(c+a)} \)

\(\\=\dfrac{a^3+a^2b-2a^2b-2ab^2+ab^2+b^3+b^3+b^2c-2b^2c-2bc^2+bc^2+c^3+c^3+c^2a-2c^a+2ca^2-ca^2+a^3}{(a+b)(b+c)(c+a)}\)

\(\\=\dfrac{(a-b)^2(a+b)+(b-c)^2(b+c)+(c-a)^2(c+a)}{(a+b)(b+c)(c+a)}\)

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=2\left(a+b+c\right)\)

\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)

\(=\dfrac{b^2-cb-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{-\left(b+a-c\right)}{ab\left(a-c\right)\left(b-c\right)}\)