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回覆於: 2022/11/23 下午 07:05:33
若x,y,z為實數 由(x^2+y^2+z^2)(1+1+1)>=(x+y+z)^2可得 3(x^2+y^2+z^2)>=81 ==>x^2+y^2+z^2>=27 但題目中之x^2+y^2+z^2=11<27 故本連立方程組x,y,z應不全為實數
另由(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx) ==>81=11+2(xy+yz+zx) ==>xy+yz+zx=35......(1) 又x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) ==>23-3xyz=9(11-35)=-216 ==>xyz=239/3......(2) 由(1),(2)及x+y+z=9可得 x,y,z為t^3-9t^2+35t-239/3=0之根 ==>3t^3-27t^2+105t-239=0 解出t即為x,y,z
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