a. để phương trình có 2 nghiệm phân biệt thì △ > 0
\(\text{△}=b^2-4ac=\left[-2\cdot\left(m-1\right)\right]^2-4\cdot1\cdot\left(4m+1\right)\\ \text{△}=4\cdot\left(m-1\right)^2-4\cdot\left(4m+1\right)\\ \text{△}=4\cdot\left[\left(m-1\right)^2-\left(4m+1\right)\right]\\ \text{△}=4\cdot\left[m^2-2m+1-4m-1\right]\\ \text{△}=4\cdot\left[m^2-m\right]=4m\cdot\left(m-6\right)\)
vậy m > 6 hoặc m < 0
b. áp dụng định lý vi-et ta có:
\(x_1+x_2=-b< =>2\left(m-1\right)\\ x_1x_2=c< =>4m+1\)
theo đề: \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=5\)
\(=>\left[2\left(m-1\right)\right]^2-2\left(4m+1\right)=5\\ 4\left(m-1\right)^2-2\left(4m+1\right)=5\\ 4\left(m^2-2m+1\right)-8m-2=5\\ 4m^2-8m+4-8m-2=5\\ 4m^2-16m-3=0\\ =>\left[{}\begin{matrix}m=\dfrac{4+\sqrt{19}}{2}\left(KTM\right)\\m=\dfrac{4-\sqrt{19}}{2}\left(TM\right)\end{matrix}\right.\)
vậy m cần tìm là: \(\dfrac{4-\sqrt{19}}{2}\)
