Lời giải:

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

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

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

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

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

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

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

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

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

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

Phân thức có nghĩa khi a;b;c không đồng thời bằng 0

Khi đó:

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

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

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

\(=a^2+b^2+c^2+ab+bc+ca\)

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

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

\(=a+b+c\)

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

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

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

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

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

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

       \(ab^2-ac^2-b^3+bc^2\)

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

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

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

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

Có a²(b-c) + b²(c-a) + c²(a-b)

= a²(b-c) - b²(a-c) + c²(a-b)

= a²(b-c) - b²(b-c+a-b) + c²(a-b)

= a²(b-c) - b²(b-c) - b²(a-b) + c²(a-b)

=[a²(b-c) - b²(b-c)] - [b²(a-b) - c²(a-b)]

=(b-c)(a²-b²) - (a-b)(b²-c²)

=(b-c)(a-b)(a+b) - (a-b)(b-c)(b+c)

=(b-c)(a-b)[(a+b)-(b+c)]

=(b-c)(a-b)(a-c)

 Có ab² - ac² - b³ + bc²

   = (ab²-ac²) - (b³-bc²)

   =a(b²-c²) - b(b²-c²)

=(b²-c²)(a-b)

=(b-c)(b+c)(a-b)

Có  a²(b-c) + b²(c-a) + c²(a-b)   /   ab² - ac² - b³ + bc²

  = (b-c)(a-b)(a-c) / (b-c)(b+c)(a-b)

= (a-c) / (b+c)

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

Tương tự:

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

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

\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)

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

cảm ơn b nha ^^

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

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

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

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

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

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

Đề lỗi rồi chứ mình ko rút gọn đc nữa

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

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

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

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