\(\text{Δ}=\left(-2\right)^2-4\cdot1\cdot\left(m-1\right)=4-4m+4=-4m+8\)

Để phương trình có hai nghiệm thì Δ>=0

=>\(-4m+8>=0\)

=>-4m>=-8

=>m<=2

Theo vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(x_1^4-x_1^3=x_2^4-x_2^3\)

=>\(x_1^4-x_2^4=x_1^3-x_2^3\)

=>\(\left(x_1-x_2\right)\left(x_1+x_2\right)\left(x_1^2+x_2^2\right)=\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1x_2\right)\)

=>\(\left(x_1-x_2\right)\cdot2\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]=\left(x_1-x_2\right)\cdot\left[\left(x_1+x_2\right)^2-x_1x_2\right]\)

=>\(\left(x_1-x_2\right)2\cdot\left[2^2-2\left(m-1\right)\right]=\left(x_1-x_2\right)\left(2^2-\left(m-1\right)\right)\)

=>\(\left(x_1-x_2\right)\cdot2\cdot\left(4-2m+2\right)=\left(x_1-x_2\right)\cdot\left(4-m+1\right)\)

=>\(\left(x_1-x_2\right)\left(12-4m\right)-\left(x_1-x_2\right)\left(5-m\right)=0\)

=>\(\left(x_1-x_2\right)\left(12-4m-5+m\right)=0\)

=>\(\left(x_1-x_2\right)\left(-3m+7\right)=0\)

=>\(\left[{}\begin{matrix}x_1-x_2=0\\-3m+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=x_2\\m=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)

Nếu \(x_1=x_2\) thì \(x_1=x_2=\dfrac{x_1+x_2}{2}=\dfrac{2}{2}=1\)

\(x_1x_2=m-1\)

=>m-1=1

=>m=2(nhận)

m=2

M=2

9

Vậy giá trị cần tìm của m cần tìm là m = 2

m = 2

m=2

Vậy giá trị m cần tìm là m=2