Đáp án: 1.$A=0$
2.$A=3$
Giải thích các bước giải:
1.Ta có:
$A=a^3+b^3+c\left(a^2+b^2\right)-abc$
$\to A=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2+b^2-ab\right)$
$\to A=\left(a+b+c\right)\left(a^2-ab+b^2\right)$
$\to A=0$ vì $a+b+c=0$
2.Ta có:
$\begin{cases}\dfrac1a+\dfrac1b+\dfrac1c=2\\ a+b+c=abc\end{cases}$
$\to\begin{cases}\dfrac1a+\dfrac1b+\dfrac1c=2\\ \dfrac1b\cdot\dfrac1c+\dfrac1c\cdot\dfrac1a+\dfrac1a\cdot \dfrac1b=1\end{cases}$
$\to A=\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2-2\left(\dfrac1b\cdot\dfrac1c+\dfrac1c\cdot\dfrac1a+\dfrac1a\cdot \dfrac1b\right)=3$
