在三角形ABC中,角A,B,C所对的边长分别是a,b,c,且(a^2+b^2-c^2)⼀(a^2-b^2+c^2)=b⼀(3a-b)

1)已知等式结合正弦、余弦定理：
(a^2+b^2-c^2)/(a^2-b^2+c^2)=b/(3a-b)
等价于：2ab·cosC/(2ac·cosB) = sinB/(3sinA-sinB)
等价于：sinB·cosC/sinC·cosB = sinB/(3sinA-sinB)
∵A、B、C是三角形内角，∴sinB>0，sinA>0，sinC>0，∴cosC/sinC·cosB = 1/(3sinA-sinB)
∴sinC·cosB = 3sinAcosC-sinBcosC
∴3sinAcosC = sinC·cosB + sinBcosC = sin(B+C) = sinA
∴cosC = 1/3，代入同角关系式，sinC = 2√2/3
2)S = (1/2)·ab·sinC = 4√2，∴ab = 12
而17 = c^2 = a^2 + b^2 - 2ab·cosC = a^2 + b^2 - (2/3)·ab
∴17 - (4/3)ab = (a-b)^2、17 + (8/3)ab = (a+b)^2
∴1 = (a-b)^2 ，49 = (a+b)^2
解得a = 4，b = 3 或者a = 3 ，b = 4