Ta có: \(a^2b+b^2c+c^2a-ab^2-bc^2-a^2c\)

\(=a^2\left(b-c\right)+a\left(c^2-b^2\right)+bc\left(b-c\right)\)

\(=\left(b-c\right)\left(a^2+bc\right)-a\left(b-c\right)\left(b+c\right)\)

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

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

\(=\left(b-c\right)\left\lbrack a\left(a-b\right)-c\left(a-b\right)\right\rbrack=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

Ta có: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)\)

\(=a^3b^2-a^3c^2+b^3c^2-a^2b^3+c^3\left(a^2-b^2\right)\)

\(=a^2b^2\left(a-b\right)-c^2\left(a^3-b^3\right)+c^3\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(a^2b^2+c^3a+c^3b\right)-c^2\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(=\left(a-b\right)\left(a^2b^2+ac^3+bc^3-a^2c^2-abc^2-c^2b^2\right)\)

\(=\left(a-b\right)\left\lbrack a^2\left(b^2-c^2\right)+ac^2\left(c-b\right)+bc^2\left(c-b\right)\right\rbrack\)

\(=\left(a-b\right)\left(b-c\right)\left\lbrack a^2\left(b+c\right)-ac^2-bc^2\right\rbrack\)

\(=\left(a-b\right)\left(b-c\right)\left\lbrack a^2b+a^2c-ac^2-bc^2\right\rbrack=\left(a-b\right)\left(b-c\right)\cdot\left\lbrack b\left(a^2-c^2\right)+ac\left(a-c\right)\right\rbrack\)

=(a-b)(b-c)(a-c)\(\left\lbrack b\left(a+c\right)+ac\right\rbrack\)

=(a-b)(b-c)(a-c)(ab+bc+ac)

Ta có: \(C=\frac{a^2b+b^2c+c^2a-ab^2-bc^2-a^2c}{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}\)

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

\(=\frac{1}{ab+bc+ac}\)

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)

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

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

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

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

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

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

Đề lỗi rồi chứ mình ko rút gọn đc nữa

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

\(=\frac{a}{b-2}-\frac{\left(a+1\right)^2}{\left(b-2\right)\left(b+2\right)}\cdot\frac{b+2}{a+1}=\frac{a}{b-2}-\frac{a+1}{b-2}\)

\(=-\frac{1}{b-2}\)

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

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

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

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

Ta có

\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)

tương tự

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

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

Cộng (1);(2) và (3) ta có

\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}=\frac{a}{a+c}-\frac{b}{a+b}+\frac{b}{a+b}-\frac{c}{b+c}+\frac{c}{c+b}-\frac{a}{a+b}=0 \)

thank bạn nha

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

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

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

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

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

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

Vậy..

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

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

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

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

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

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

\(=a+b+c\)