Bài này lớp 9 nhé ạ vietttt

Giải thích các bước giải:

Để phương trình có hai nghiệm phân biệt

$\to \Delta>0$

$\to (2m-1)^2-4(m^2-7)> 0$ 

$\to -4m+29> 0$

$\to m<\dfrac{29}4$

$\to \begin{cases}x_1+x_2=2m-1\\x_1x_2=m^2-7\end{cases}$

Ta có:

$4x_1^2-x_1-3x_2^2+x_2=x_1x_2$

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

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

$\to x_1^2+3(x_1+x_2)(x_1-x_2)-(x_1-x_2)=x_1x_2$

$\to x_1^2+3(2m-1)(x_1-x_2)-(x_1-x_2)=x_1x_2$

$\to x_1^2+(3(2m-1)-1)(x_1-x_2)=m^2-7$

$\to x_1^2+(6m-4)(x_1-x_2)=m^2-7$

Mà $x_1^2-(2m-1)x_1+m^2-7=0$

$\to x_1^2=(2m-1)x_1-m^2+7$

$\to (2m-1)x_1-m^2+7+(6m-4)(x_1-x_2)=m^2-7$

$\to (2m-1)x_1-m^2+7+(6m-4)(x_1-(2m-1-x_1))=m^2-7$

$\to (2m-1)x_1+(6m-4)(x_1-(2m-1-x_1))=2m^2-14$

$\to  14mx_1-9x_1-12m^2+14m-4=2m^2-14$

$\to x_1\left(14m-9\right)=14m^2-14m-10$

$\to x_1=\dfrac{14m^2-14m-10}{14m-9}$

$\to (\dfrac{14m^2-14m-10}{14m-9})^2-(2m-1)\cdot \dfrac{14m^2-14m-10}{14m-9}+m^2-7=0$

$\to \left(14m^2-14m-10\right)^2-\left(14m^2-14m-10\right)\left(2m-1\right)\left(14m-9\right)+m^2\left(14m-9\right)^2-7\left(14m-9\right)^2=0$

$\to \left(m-1\right)\left(4m-29\right)\left(49m-13\right)=0$

$\to 196m^4-392m^3-84m^2+280m+100-392m^4+840m^3-294m^2-194m+90+196m^4-252m^3+81m^2-1372m^2+1764m-567=0$

$\to 196m^3-1669m^2+1850m-377=0$

$\to m\in\{1, \dfrac{29}4, \dfrac{13}{49}$