Giải thích các bước giải:
Trường hợp hai đa thức có nghiệm chung $x=0$
$\to c=c_1=0$
$\to (ac_1-a_1c)^2=(ab_1-a_1b)(bc_1-b_1c)(=0)$
$\to đpcm$
Trường hợp hai đa thức có nghiệm chung $x\ne 0$
Giả sử hai đa thức có nghiệm chung $x=x_0, x_0\ne 0$
$\to \begin{cases}ax_0^2+bx_0+c=0\\a_1x_0^2+b_1x_0+c_1=0\end{cases}$
$\to \begin{cases}ac_1x_0^2+bc_1x_0+cc_1=0\\a_1cx_0^2+b_1cx_0+c_1c=0\end{cases}$
$\to (ac_1x_0^2+bc_1x_0+cc_1)-(a_1cx_0^2+b_1cx_0+c_1c)=0$
$\to (ac_1-a_1c)x_0^2+(bc_1-b_1c)x_0=0$
$\to (ac_1-a_1c)x_0^2=-(bc_1-b_1c)x_0$
$\to (ac_1-a_1c)x_0=-(bc_1-b_1c)$
$\to (ac_1-a_1c)^2x^2_0=(bc_1-b_1c)^2$
Ta có:
$\begin{cases}ax_0^2+bx_0+c=0\\a_1x_0^2+b_1x_0+c_1=0\end{cases}$
$\begin{cases}ab_1x_0^2+bb_1x_0+cb_1=0\\a_1bx_0^2+b_1bx_0+bc_1=0\end{cases}$
$\to (ab_1x_0^2+bb_1x_0+cb_1)-(a_1bx_0^2+b_1bx_0+bc_1)=0$
$\to (ab_1-a_1b)x_0^2+(cb_1-bc_1)=0$
$\to x_0^2=-\dfrac{cb_1-bc_1}{ab_1-a_1b}$
$\to x_0^2=\dfrac{bc_1-cb_1}{ab_1-a_1b}$
$\to (ac_1-a_1c)^2\cdot \dfrac{bc_1-cb_1}{ab_1-a_1b}=(bc_1-b_1c)^2$
$\to (ac_1-a_1c)^2=(bc_1-cb_1)(ab_1-a_1b)$
$\to đpcm$
