Bài này nữa thầy ơi :Cho pt x^2 -(m-2)x-3=0( m là tham số ).Chứng minh pt luôn có hai nghiệm phân biệt .Tìm m để nghiệm đó thỏa mãn hệ thức căn (x1^2 +2019) -x1= căn(x2^2+2019) +x2

$$\eqalign{ & {x^2} - \left( {m - 2} \right)x - 3 = 0 \cr & \Delta = {\left( {m - 2} \right)^2} + 4.1.\left( { - 3} \right) \cr & = {\left( {m - 2} \right)^2} + 12 > 0\,\,\forall m \in R \cr & \Rightarrow PT\,\,luon\,\,co\,\,2\,\,nghiem\,\,phan\,\,biet. \cr & Ap\,\,dung\,\,DL\,\,Vi - et:\,\,\left\{ \matrix{ {x_1} + {x_2} = m - 2 \hfill \cr {x_1}{x_2} = - 3 \hfill \cr} \right. \cr & Theo\,\,gt: \cr & \sqrt {x_1^2 + 2019} - {x_1}{\kern 1pt} = \sqrt {x_2^2 + 2019} + {x_2} \cr & \Leftrightarrow \sqrt {x_1^2 + 2019} - \sqrt {x_2^2 + 2019} = {x_1} + {x_2} \cr & \Leftrightarrow {{x_1^2 + 2019 - x_2^2 - 2019} \over {\sqrt {x_1^2 + 2019} + \sqrt {x_2^2 + 2019} }} = {x_1} + {x_2} \cr & \Leftrightarrow {{\left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right)} \over {\sqrt {x_1^2 + 2019} + \sqrt {x_2^2 + 2019} }} = {x_1} + {x_2} \cr & \Leftrightarrow \left( {{x_1} + {x_2}} \right)\left[ {{{{x_1} - {x_2}} \over {\sqrt {x_1^2 + 2019} + \sqrt {x_2^2 + 2019} }} - 1} \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ {x_1} + {x_2} = 0\,\,\left( 1 \right) \hfill \cr {{{x_1} - {x_2}} \over {\sqrt {x_1^2 + 2019} + \sqrt {x_2^2 + 2019} }} = 1\,\,\left( 2 \right) \hfill \cr} \right. \cr & \left( 1 \right) \Leftrightarrow m - 2 = 0 \Leftrightarrow m = 2 \cr & \left( 2 \right) \Leftrightarrow \sqrt {x_1^2 + 2019} + \sqrt {x_2^2 + 2019} = {x_1} - {x_2} \cr & \Leftrightarrow \left\{ \matrix{ {x_1} - {x_2} \ge 0 \hfill \cr x_1^2 + 2019 + x_2^2 + 2019 + 2\sqrt {\left( {x_1^2 + 2019} \right)\left( {x_2^2 + 2019} \right)} = x_1^2 + x_2^2 - 2{x_1}{x_2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {x_1} \ge {x_2} \hfill \cr 2\sqrt {\left( {x_1^2 + 2019} \right)\left( {x_2^2 + 2019} \right)} + 4038 + 2{x_1}{x_2} = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {x_1} \ge {x_2} \hfill \cr 2\sqrt {\left( {x_1^2 + 2019} \right)\left( {x_2^2 + 2019} \right)} + 4038 - 6 = 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ {x_1} \ge {x_2} \hfill \cr 2\sqrt {\left( {x_1^2 + 2019} \right)\left( {x_2^2 + 2019} \right)} + 4032 = 0\,\left( {Vo\,\,nghiem} \right) \hfill \cr} \right. \cr & Vay\,m = 2. \cr} $$