a: ĐKXĐ: \(\left(a+b+c\right)^2-\left(ab+bc+ca\right)<>0\)

=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)-\left(ab+ac+bc\right)<>0\)

=>\(a^2+b^2+c^2+ab+ac+bc<>0\)

=>\(2a^2+2b^2+2c^2+2\left(ab+ac+bc\right)<>0\)

=>\(\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)<>0\)

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

Dấu '=' xảy ra khi \(\begin{cases}a+b=0\\ b+c=0\\ a+c=0\end{cases}\Rightarrow a=b=c=0\)

=>Để M xác định thì \(a^2+b^2+c^2<>0\)

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

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

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

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

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

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

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

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

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

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

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

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

Khi đó:

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

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

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

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

đây là một hằng đẳng thức nha bạn

=a³+b³+c³-3abc

thank

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

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

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

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

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

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

\(=a+b+c\)