Bài 9:
${{x}^{2}}-2\left( a-1 \right)x+2a-5=0$
a)
$\Delta '=b{{'}^{2}}-ac$
$\Delta '={{\left[ -\left( a-1 \right) \right]}^{2}}-1\left( 2a-5 \right)$
$\Delta '={{a}^{2}}-2a+1-2a+6$
$\Delta '={{a}^{2}}+7>0\,$với mọi $a\in \mathbb{R}$
Vậy phương trình luôn có nghiệm với mọi $a$
b)
Theo hệ thức Vi – et, ta có:
$\begin{cases}x_1+x_2=2a-2\\\\x_1x_2=2a-5\end{cases}$
${{x}_{1}}\,\,<\,\,1\,\,\to \,\,{{x}_{1}}-1\,\,<\,\,0$
${{x}_{2}}\,\,>\,\,1\,\,\to \,\,{{x}_{2}}-1\,\,>\,\,0$
$\to \left( {{x}_{1}}-1 \right)\left( {{x}_{2}}-1 \right)<0$
$\to {{x}_{1}}{{x}_{2}}-\left( {{x}_{1}}+{{x}_{2}} \right)+1\,\,<\,\,0$
$\to 2a-5\,-\,\left( 2a-2 \right)+1<0$
$\to 2a\,-\,5\,-\,2a\,+2\,+\,1\,\,<\,\,0$
$\to -2\,\,<\,\,0$ ( luôn đúng )
Vậy $a\in \mathbb{R}$thì ${{x}_{1}}\,\,<\,\,1\,\,<\,\,{{x}_{2}}$
c)
$\,\,\,\,\,\,{{x}_{1}}^{2}+{{x}_{2}}^{2}=6$
$\to {{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}-2{{x}_{1}}{{x}_{2}}=6$
$\to {{\left( 2a-2 \right)}^{2}}-2\left( 2a-5 \right)=6$
$\to 4{{a}^{2}}-8a+4-4a+10=6$
$\to 4{{a}^{2}}-12a+8=0$
$\to {{a}^{2}}-3a+2=0$
$\to {{a}^{2}}-2a-a+2=0$
$\to a\left( a-2 \right)-\left( a-2 \right)=0$
$\to \left( a-2 \right)\left( a-1 \right)=0$
$\to\left[\begin{array}{1}a=2\\\\a=1\end{array}\right.$
Vậy $a=2$ hoặc $a=1$ là giá trị cần tìm
