`8x^2-8x+m^2+1=0`
`\Delta'=(-4)^2-(m^2+1).8`
`\Delta'=16-8m^2-8`
`\Delta'=-8m^2+8`
Để pt có nghiệm thì
`-8m^2+8>=0`
`<=> -8m^2>=-8`
`<=> m^2<=1`
`<=> -1<=m<=1`
Theo Viet: $\begin{cases}x_1+x_2=1 (1)\\x_1.x_2=\dfrac{m^2+1}{8} (2)\end{cases}$
Có: `(x_1)^4-(x_2)^4=(x_1)^3-(x_2)^3`
`<=> (x_1)^4-(x_2)^4-(x_1)^3+(x_2)^3=0`
`<=> (x_1)^3.(x_1-1)-(x_2)^3(x_2-1)=0`
`-> (x_1)^3.(x_1-x_1-x_2)-(x_2)^3.(x_2-x_1-x_2)=0`
`<=> (x_1)^3.(-x_2)-(x_2)^3.(-x_1)=0`
`<=> -x_1.x_2 (x_1^2-x_2^2)=0`
`<=> -x_1.x_2.(x_1-x_2)(x_1+x_2)=0`
`-> -(m^2+1)/8. 1.(x_1-x_2)=0`
`<=> x_1-x_2=0`
`<=> x_1=x_2`
`(2)-> x_1=x_2=(\sqrt{2m^2+2})/4`
Thay `x_1=x_2=(\sqrt{2m^2+2})/4` vào (1) ta có:
`(\sqrt{2m^2+2})/2=1`
`<=> \sqrt{2m^2+2}=2`
`<=> 2m^2+2=4`
`<=> m^2=1`
`<=> m=+-1` (TM)
Vậy `m=+-1`
