giúp mik giải bài này vs ạ

Giải thích các bước giải:

Để phương trình có $2$ nghiệm phân biệt

$\to \Delta'>0$

$\to (-2)^2-1\cdot (m^2+2)>0$ 

$\to 2-m^2>0$

$\to m^2<2$

$\to -\sqrt2<m<\sqrt2$

$\to \begin{cases}x_1+x_2=4\\x_1x_2=m^2+2\end{cases}$

Ta có:
$\dfrac{x_1}{x_2}+\dfrac{x_1x_2+x_2^2}{2x_1}+\dfrac{x_2-x_1}{m^2+2}=3$

$\to \dfrac{x_1}{x_2}+\dfrac{x_2(x_1+x_2)}{2x_1}+\dfrac{x_2-x_1}{x_1x_2}=3$

$\to \dfrac{x_1}{x_2}+\dfrac{4x_2}{2x_1}+\dfrac{x_2-x_1}{x_1x_2}=3$

$\to \dfrac{x_1}{x_2}+\dfrac{2x_2}{x_1}+\dfrac{x_2-x_1}{x_1x_2}=3$

$\to \dfrac{x_1^2+2x_2^2+x_2-x_1}{x_1x_2}=3$

$\to \dfrac{(4-x_2)^2+2x_2^2+x_2-(4-x_2)}{x_1x_2}=3$

$\to \dfrac{3x_2^2-6x_2+12}{x_1x_2}=3$

$\to \dfrac{x_2^2-2x_2+4}{x_1x_2}=1$

$\to \dfrac{4x_2-m^2-2-2x_2+4}{x_1x_2}=1$

$\to \dfrac{2x_2-m^2+2 }{x_1x_2}=1$

$\to 2x_2-m^2+2=x_1x_2$

$\to 2x_2-m^2+2=m^2+2$

$\to 2x_2=2m^2$

$\to x_2=m^2$

$\to (m^2)^2-4m^2+m^2+2=0$

$\to m^4-3m^2+2=0$

$\to (m^2-2)(m^2-1)=0$

$\to m=\pm\sqrt2, m=\pm1$

$\to m=\pm1$ vì $-\sqrt2<m<\sqrt2$