Question

va rog dau 70 de puncte raspundeti

Answer 1

$a)\ m=3 \Rightarrow 2x^2-3x+1=0 \Rightarrow 2x^2-x-2x+1=0\Rightarrow x(2x-1)-(2x-1)=0 \Rightarrow \\ \\ \Rightarrow (2x-1)(x-1)=0 \Rightarrow \begin{cases}\it2x-1=0 \Rightarrow x_1=\dfrac{1}{2}\\ \\ \itx-1=0 \Rightarrow x_2=1\end{cases} \Rightarrow S=\Big\{\dfrac{1}{2},\ 1\Big\}$

$\it b)\ x_1,x_2\in\mathbb{R} \Rightarrow \Delta\geq0 \Rightarrow (-3)^2-8(m-2)\geq0 \Rightarrow 9\geq8m-16 \Rightarrow \\ \\ \Rightarrow 8m\leq25 \Rightarrow m\leq\dfrac{25}{8} \Rightarrow m\in\Big(-\infty,\ \dfrac{25}{8}\Big]$

$\it c)\ x=m-\dfrac{1}{2} \Rightarrow 2(m-\dfrac{1}{2})^2-3(m-\dfrac{1}{2})+m-2=0 \Rightarrow \\ \\ \\ \Rightarrow (m-\dfrac{1}{2})(2m-1-3)+m-2=0 \Rightarrow \dfrac{2m-1}{2}(2m-4)+m-2=0 \Rightarrow \\ \\ \Rightarrow \dfrac{2m-1}{2}(m-2)\cdot2+m-2=0 \Rightarrow (m-2)(2m-1+1)=0\Rightarrow\\ \\ \dfrac{2m-1}{2}(2m-4)+m-2=0 \Rightarrow \dfrac{2m-1}{2}(m-2)\cdot2+(m-2)=0 \Rightarrow \\ \\ \Rightarrow (m-2)(2m-1+1)=0 \Rightarrow (m-2)\cdot2m=0 \Rightarrow m\in\{0,\ 2\}$