`color{#bfbfbf}{#}color{#858585}{d}color{#424242}{t}color{#050505}{t}`

`x^2-mx+1=0`

`\Delta=(-m)^2-4=m^2-4`

Để pt có `2` nghiệm thì `\Delta>=0`

`-> m^2-4>=0`

`-> (m-2)(m+2)>=0`

TH1: `{(m-2>=0),(m+2>=0):}`

`-> {(m>=2),(m>=-2):} -> m>=2`

TH2: `{(m-2<=0),(m+2<=0):}`

`-> {(m<=2),(m<=-2):}->m<=-2`

Vậy `m<=-2` và `m>=2` thì pt có `2` nghiệm `x_1,x_2`

`{(x_1+x_2=m),(x_1x_2=1):}` (Viète)

Theo đề:

`1/(\sqrt(x_1^2+1)+x_1)=2\sqrt(2)-x_1-\sqrt(x_2^2+1)`

`(\sqrt(x_1^2+1)-x_1)/(x_1^2+1-x_1^2)=2\sqrt(2)-x_1-\sqrt(x_2^2+1)`

`\sqrt(x_1^2+1)-x_1=2\sqrt(2)-x_1-\sqrt(x_2^2+1)`

`\sqrt(x_1^2+1)+\sqrt(x_2^2+1)=2\sqrt(2)-x_1+x_1`

`\sqrt(x_1^2+x_1x_2)+\sqrt(x_2^2+x_1x_2)=2\sqrt(2)`

`\sqrt(x_1(x_1+x_2))+\sqrt(x_2(x_1+x_2))=2\sqrt(2)`

`\sqrt(x_1m)+\sqrt(x_2m)=2\sqrt(2)`

`x_1m+2\sqrt(x_1x_2m^2)+x_2m=8`

`m(x_1+x_2)+2\sqrt(m^2)=8`

`m^2+2|m|=8`

`m^2+2|m|-8=0`

`TH1: m^2-2m-8=0`

`-> m^2+2m-4m-8=0`

`-> m(m+2)-4(m+2)=0`

`-> (m-4)(m+2)=0`

`-> m=4\ (tm)` hoặc `m=-2\ (tm)`

`TH2: m^2+2m-8=0`

`-> m^2+4m-2m-8=0`

`-> m(m+4)-2(m+4)=0`

`-> (m-2)(m+4)=0`

`-> m=2\ (tm)` hoặc `m=-4\ (tm)`

Vậy `m\in {+-2;+-4}`