(a+b)*(a^2-ab+b^2)+(a-b)*(a^2+ab+b^2)

=a^3+b^3+a^3-b^3

=2a^3

tks cho mk nhe

đây là một hằng đẳng thức nha bạn

=a³+b³+c³-3abc

thank

cho mình cách giải

Với a + b + c = 0 , ta có :

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

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

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

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

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

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

\(P=\dfrac{a^2}{ab+b^2}+\dfrac{b^2}{ab-a^2}-\dfrac{a^2+b^2}{ab}\) (\(a\ne b;a\ne0;a\ne-b;b\ne0\))

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

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

\(=\dfrac{a^4-a^3b-b^3a-b^4-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a+b\right)\left(a-b\right)}\)

\(=\dfrac{a^4-a^3b-b^3a-b^4-\left(a^4-b^4\right)}{ab\left(a+b\right)\left(a-b\right)}\)

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

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

b) -Ta có: \(P=0\)

\(\Leftrightarrow-\dfrac{a^2+b^2}{a^2-b^2}=0\)

\(\Leftrightarrow a^2+b^2=0\)

-Vì \(a^2\ge0;b^2\ge0\)

\(\Rightarrow a=0;b=0\) (không thỏa mãn điều kiện).

-Vậy không có giá trị nào của a,b để \(P=0\).

c) 

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

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

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

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

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

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

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

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

Đề lỗi rồi chứ mình ko rút gọn đc nữa