Sửa đề:

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

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

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

\(=a+b+c\left(a^2+b^2+c^2+ab+bc+ca\ne0\right)\)

cảm ơn anh để em xem lại 

(a^2+b^2+c^2)(a+b+c)+(ab+ac+bc)^2/(a+b+c)^2-(ab+bc+ca) có đúng đề ko ạ 

không biết làm hâhha

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

\(A=x^3+6x^2-7x-60\)

Nếu rút gọn thành nhân tử thì:

\(A=x^3-3x^2+9x^2-27x+20x-60=x^2\left(x-3\right)+9x\left(x-3\right)+20\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+9x+20\right)=\left(x-3\right)\left(x^2+4x+5x+20\right)=\left(x-3\right)\left[x\left(x+4\right)+5\left(x+4\right)\right]\)

\(A=\left(x-3\right)\left(x+4\right)\left(x+5\right)\).

đây là một hằng đẳng thức nha bạn

=a³+b³+c³-3abc

thank

cho mình cách giải

Với a + b + c = 0 , ta có :

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

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

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

\(\Leftrightarrow A=\frac{ab}{-2ab}\)\(+\frac{bc}{-2bc}\)\(+\frac{ac}{-2ac}\)

\(\Leftrightarrow A=\frac{-1}{2}\)\(+\frac{-1}{2}\)\(+\frac{-1}{2}\)

\(\Leftrightarrow A=\frac{-3}{2}\)

      \(\frac{a}{ab}+a+1+\frac{b}{bc}+b+1+\frac{c}{ca}+c+1\)

\(=\frac{1}{b}+a+1+\frac{1}{c}+b+1+\frac{1}{c}+c+1\)

\(=3+a+b+c+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}\)

\(=3+\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)

\(...............................................................\)

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

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

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

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

- Phân tích ra nhân tử :

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

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

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

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

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

Từ đây ta có \(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(\Rightarrow A=a+b+c\)