求方程组实数解5(x+1/x)=12(y+1/y)=13(z+1/z),xy+yz+zx=1提示：可用三角换元也可死算

Question

求方程组实数解5(x+1/x)=12(y+1/y)=13(z+1/z),xy+yz+zx=1 提示：可用三角换元也可死算

Answer 1

求方程组实数解5(x+1/x)=12(y+1/y)=13(z+1/z),xy+yz+zx=1
由xy+yz+zx=1,得1/x+1/y+1/z=1/xyz.(1)
由(1/x+1/y+1/z)²=1/x²+1/y²+1/z²+2(1/xy+1/yz+1/zx)=1/x²+1/y²+1/z²+2(x+y+z)/xyz=1/(xyz)²
得1/x²+1/y²+1/z²=1/(xyz)²-2(x+y+z)/xyz.(2)
由(x+y+z)²=x²+y²+z²+2(xy+yz+zx)=x²+y²+z²+2
得x²+y²+z²=(x+y+z)²-2.(3)
令5(x+1/x)=12(y+1/y)=13(z+1/z)=m,则
x+1/x=m/5.(a) y+1/y=m/12 .(b) z+1/z=m/13.(c)
(a)²+(b)²+(c)²得;
x²+y²+z²+6+1/x²+1/y²+1/z²=m²(1/25+1/144+1/169)=32161m²/608400
将(2)(3)代入得:
(x+y+z)²-2(x+y+z)/xyz+1/(xyz)²=32161m²/608400-4
即有 [(x+y+z)-1/xyz]²=32161m²/608400-4
于是得 (x+y+z)-1/xyz=√(32161m²/608400-4).(4)
(a)+(b)+(c)得:
(x+y+z)+(1/x+1/y+1/z)=m(1/5+1/12+1/13)=281m/780
将(1)代入得:
(x+y+z)+1/xyz=281m/780.(5)
由(4)+(5)得;
x+y+z=(1/2)[281m/780+√(32161m²/608400-4)].(6)
(5)-(4)得:
1/xyz=(1/2)[281m/780-√(32161m²/608400-4)].(7)