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

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

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

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

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

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

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

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

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

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

\(P=0\)

Xét: \(f\left(x\right)=\frac{x^2-bc}{\left(x+b\right)\left(x+c\right)}+\frac{b^2-xc}{\left(b+c\right)\left(b+x\right)}+\frac{c^2-xb}{\left(c+x\right)\left(c+b\right)}\)

\(\Rightarrow f\left(a\right)=P\)

Ta có: \(f\left(b\right)=\frac{b^2-bc}{2b\left(b+c\right)}+\frac{b^2-bc}{2b\left(b+c\right)}+\frac{c^2-b^2}{\left(c+b\right)\left(c+b\right)}\)

\(\Rightarrow f\left(b\right)=\frac{2b\left(b-c\right)}{2b\left(b+c\right)}+\frac{\left(c-b\right)\left(c+b\right)}{\left(c+b\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0\left(1\right)\)

Chứng minh tương tự ta cũng có: \(f\left(c\right)=0\left(2\right)\)

Từ (1) và (2) suy ra \(f\left(x\right)=0\left(\forall x\right)\Rightarrow f\left(a\right)=0\left(\forall x\right)\)

Vậy A =0

mình ghi sai rồi \(P=1\)mới đúng chứ kg phải A =1

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

thiếu đề bài òi bạn ko làm đc đâu

Mik giải ra rồi!

giải kiểu gì ??????

😥 😥 😥

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

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