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

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

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

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

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

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

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

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

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

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 ^^

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

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

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

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

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

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

       \(ab^2-ac^2-b^3+bc^2\)

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

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

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)

quy đồng là ra

Cho phân thức \(A=\frac{x^5+2x^4+2x^3-4x^2+3x+6}{x^2+2x-8}\)

a) Tìm tập  xác định của A

b) Tìm các giá trị của x để A = 0

c) Rút gọn A

giải giúp

a, Đk để phân thức M có nghĩa là mẫu khác 0

Xét: \(\left(a+b+c\right)^2-\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+ab+bc+ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)

\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2=0\)

\(\Leftrightarrow a+b=b+c=a+c=0\)

\(\Leftrightarrow a=b=c\)

Vậy để M có nghĩa thì \(a^2+b^2+c^2\ne0\)

b, Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)

Đặt: \(\hept{\begin{cases}a^2+b^2+c^2=x\\ab+bc+ca=y\end{cases}}\)

Khi đó ta được: \(\left(a+b+c\right)^2=x+2y\)

Ta có: \(M=\frac{x\left(x+2y\right)+y^2}{x+2y-y}=\frac{x^2+2xy+y^2}{x+y}=\frac{\left(x+y\right)^2}{x+y}=x+y\)

\(=a^2+b^2+c^2+ab+bc+ac\left(Đkxđ:a^2+b^2+c^2\ne0\right)\)

Thay \(ab+bc+ca=1\) ta có:

\(1+a^2=ab+bc+ca+a^2=b\left(c+a\right)+a\left(c+a\right)=\left(c+a\right)\left(a+b\right)\)

Tương tự: \(1+b^2=\left(b+c\right)\left(a+b\right);\) \(1+c^2=\left(c+a\right)\left(b+c\right)\)

\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

\(\Rightarrow\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}=1\). Vậy biểu thức đó rút gọn lại bằng 1.