Question

Biết \(\left\{{}\begin{matrix}x_1+x_2=3m\\x_1x_2=3m-2\end{matrix}\right.\). Tính \(x_1^2-3mx_2-m+1=8\)

Answer 1

Lời giải:

$A=x_1^2-3mx_2-m+1=x_1^2-(x_1+x_2)x_2-m+1$

$=x_1^2-x_2^2-x_1x_2-m+1$

$=(x_1-x_2)(x_1+x_2)-(3m-2)-m+1$

$=3m(x_1-x_2)-4m+3$

Nếu $x_1\geq x_2$ thì:

$A=3m\sqrt{(x_1-x_2)^2}-4m+3$

$=3m\sqrt{(x_1+x_2)^2-4x_1x_2}-4m+3$

$=3m\sqrt{9m^2-4(3m-2)}-4m+3$

$=3m\sqrt{9m^2-12m+8}-4m+3$
Nếu $x_1<x_2$ thì:

$A=-3m(x_2-x_1)-4m+3$

$=-3m\sqrt{(x_1-x_2)^2}-4m+3$

$=-3m\sqrt{(x_1+x_2)^2-4x_1x_2}-4m+3$

$=-3m\sqrt{9m^2-4(3m-2)}-4m+3$

$=-3m\sqrt{9m^2-12m+8}-4m+3$