`\qquad (x^2-3x+1)^3+(x^2-5x+2)^3-(2x^2-8x+3)^3=0`
Đặt `x^2-3x+1=a; x^2-5x+2=b`
`-> 2x^2-8x+3=a+b`
Khi đó ta có pt ẩn phụ `a;b` là:
`\qquad a^3+b^3-(a+b)^3=0`
`<=> a^3+b^3-[a^3+3ab(a+b)+b^3]=0`
`<=> -3ab(a+b)=0`
`<=> [(a=0),(b=0),(a+b=0):}`
`=> [(x^2-3x+1=0(1)),(x^2-5x+2=0(2)),(2x^2-8x+3=0(3)):}`
Giải pt (1): `x^2-3x+1=0`
`<=> x^2-2.x. 3/2+9/4-5/4=0`
`<=> (x-3/2)^2=5/4`
`<=>`\(\left[ \begin{array}{l}x-\dfrac{3}{2}=\dfrac{\sqrt{3}}{2}\\x-\dfrac{3}{2}=-\dfrac{\sqrt{3}}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{3+\sqrt{3}}{2}\\x=\dfrac{3-\sqrt{3}}{2}\end{array} \right.\)
Giải pt (2): `x^2-5x+2=0`
`<=> x^2-2.x. 5/2+25/4-17/4=0`
`<=> (x-5/2)^2=17/4`
`<=>`\(\left[ \begin{array}{l}x-\dfrac{5}{2}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{17}}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{5-\sqrt{17}}{2}\end{array} \right.\)
Giải pt (3): `2x^2-8x+3=0`
`<=> x^2-4x+3/2=0`
`<=> (x^2-4x+4)-5/2=0`
`<=> (x-2)^2=5/2`
`<=>`\(\left[ \begin{array}{l}x-2=\dfrac{\sqrt{10}}{2}\\x-2=-\dfrac{\sqrt{10}}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{4+\sqrt{10}}{2}\\x=\dfrac{4-\sqrt{10}}{2}\end{array} \right.\)
Vậy `S={(3+-sqrt{3})/2;(5+-sqrt{17})/2;(4+-sqrt{10})/2}`
(Mình xin phép làm khác yêu cầu đề chút vì áp dụng HĐT kia lâu hơn ;;-;;)