$\left\{\begin{matrix} a+b+c=0\\ \alpha+\beta+\gamma=0\\ \dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}=0 \end{matrix}\right.$

Bài toán:

Cho các số $a;b;c;\alpha;\beta;\gamma$ thỏa mãn hệ:

$\left\{\begin{matrix} a+b+c=0\\ \alpha+\beta+\gamma=0\\ \dfrac{\alpha}{a}+\dfrac{\beta}{b}+\dfrac{\gamma}{c}=0 \end{matrix}\right.$

Tính giá trị biểu thức $P=\alpha a^2+\beta b^2 +\gamma c^2$

Lời giải:

ĐK: $a;b;c\neq 0$

Có: 

+) $PT1\Rightarrow c=-(a+b)$

+) $PT2\Rightarrow \gamma =-(\alpha +\beta )$

+) $PT3\Rightarrow \alpha .bc+\beta .ca+\gamma .ab=\alpha .b.\left [-(a+b)  \right ]+\beta .a.\left [-(a+b)  \right ]+\left [ -(\alpha +\beta ) \right ].ab=0$
$\Leftrightarrow (a+b)(\alpha b+\beta a)+ab(\alpha +\beta )=0$
$\Leftrightarrow \alpha b^2+\beta a^2+2ab(\alpha +\beta )=0$

$\Leftrightarrow \alpha b^2+\beta a^2-2ab\gamma =0$

Cmtt ta có: $\beta c^2+\gamma b^2-2bc\alpha =0;\gamma a^2+\alpha c^2-2ca\beta =0$

Cộng 3 đẳng thức trên lại ta có:

$(\alpha +\beta )c^2+(\gamma +\alpha )b^2+(\beta +\gamma )a^2-2(bc\alpha +ca\beta +ab\gamma )=0$
​$\Leftrightarrow \alpha a^2+\beta b^2+\gamma c^2+2(bc\alpha +ca\beta +ab\gamma )=0$
$\Leftrightarrow \alpha a^2+\beta b^2+\gamma c^2+2abc\left (\dfrac{\alpha }{a}+\dfrac{\beta }{b}+\dfrac{\gamma c}{c}  \right )=0$
$\Leftrightarrow P=\alpha a^2+\beta b^2 +\gamma c^2=0$ (đpcm)