a/ Pt hoành độ giao điểm
\(x^2=2(a-1)x+5-2a\\↔x^2-2(a-1)x+2a-5=0\\a=2→x^2-2x-1=0\\↔(x-1)^2-2=0\\↔(x-1-\sqrt 2)(x-1+\sqrt 2)=0\\↔\left[\begin{array}{1}x-1-\sqrt 2=0\\x-1+\sqrt 2=0\end{array}\right.\\↔\left[\begin{array}{1}x=1+\sqrt 2\\x=1-\sqrt 2\end{array}\right.\)
b/ \(Δ'=[-(a-1)]^2-1.(2a-5)=a^2-2a+1-2a+5=a^2-4a+4+2=(a-2)^2+2>0∀a\)
→ Hai hàm số cắt nhau tại 2 điểm phân biệt ∀a
Theo Vi-ét:
\(\begin{cases}x_1+x_2=2(a-1)\\x_1x_2=2a-5\end{cases}\)
\(x_1^2+x_2^2=6\\↔x_1^2+2x_1x_2+x_2^2-2x_1x_2=6\\↔(x_1+x_2)^2-2x_1x_2=6\\↔[2(a-1)]^2-2(2a-5)=6\\↔4a^2-8a+4-4a+10-6=0\\↔4a^2-12a+8=0\\↔a^2-3a+2=0\\↔a^2-2a-a+2=0\\↔a(a-2)-(a-2)=0\\↔(a-1)(a-2)=0\\↔\left[\begin{array}{1}a-1=0\\a-2=0\end{array}\right.\\↔\left[\begin{array}{1}a=1\\a=2\end{array}\right.\)
Vậy \(a=1\) hoặc \(a=2\)
