Đặt \(x^2=t\) \(\Rightarrow t^2+\left(1-m\right)t+2m-2=0\) (1)

Pt đã cho có 4 nghiệm pb \(\Leftrightarrow\) (1) có 2 nghiệm dương pb

\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(1-m\right)^2-8\left(m-1\right)>0\\t_1+t_2=m-1>0\\t_1t_2=2m-2>0\end{matrix}\right.\) \(\Rightarrow m>9\)

Khi đó, do vai trò của \(x_1;x_2;x_3;x_4\) như nhau, ko mất tính tổng quát, giả sử \(x_1=-\sqrt{t_1};x_2=\sqrt{t_1}\) ; \(x_3=-\sqrt{t_2};x_4=\sqrt{t_2}\)

\(\Rightarrow x_1x_2x_3x_4=t_1t_2\) ; \(x_1^2=x_2^2=t_1\) ; \(x_3^2=x_4^2=t_2\)

\(\Rightarrow\dfrac{x_1x_2x_3x_4}{2x_4^2}+\dfrac{x_1x_2x_3x_4}{2x_3^2}+\dfrac{x_1x_2x_3x_4}{2x_2^2}+\dfrac{x_1x_2x_3x_4}{2x_1^2}=2017\)

\(\Leftrightarrow\dfrac{t_1t_2}{2t_2}+\dfrac{t_1t_2}{2t_2}+\dfrac{t_1t_2}{2t_1}+\dfrac{t_1t_2}{2t_1}=2017\)

\(\Leftrightarrow t_1+t_2=2017\)

\(\Leftrightarrow m-1=2017\Rightarrow m=2018\)

a.

Ta co:

\(\orbr{\begin{cases}x^2-2x-3=0\left(1\right)\left(x\ge0\right)\\x^2+2x-3=0\left(2\right)\left(x< 0\right)\end{cases}}\)

(1)\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(l\right)\\x=3\left(n\right)\end{cases}}\)

(2)\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=-3\left(n\right)\end{cases}}\)

b.

Ta lai co:

\(\orbr{\begin{cases}x^2-2x+1-4a^2=0\left(3\right)\left(x\ge0\right)\\x^2+2x+1-4a^2=0\left(4\right)\left(x< 0\right)\end{cases}}\)

Xet (3)

De phuong trinh dau co 4 nghiem thi PT(3) co nghiem

\(\Rightarrow\Delta^`>0\)

\(\Leftrightarrow4a^2>0\)

\(\Leftrightarrow a>0\)

\(\Rightarrow x_1=1+2a;x_2=1-2a\)

Tuong tu

(4)

\(a>0\)

\(\Rightarrow x_3=-1+2a;x_4=-1-2a\)

\(\Rightarrow S=\left(1+2a\right)^2+\left(1-2a\right)^2+\left(-1+2a\right)^2+\left(-1-2a\right)^2\)

\(=2\left(1+2a\right)^2+2\left(1-2a\right)^2\)

\(\Rightarrow S< +\infty\)