若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
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若x^2-3x+1=0y^2-3y+1=0且x不等于y求x^4+y^4的值若x^2-3x+1=0y^2-3y+1=0且x不等于y求x^4+y^4的值若x^2-3x+1=0y^2-3y+1=0且x不等于
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
若x^2-3x+1=0 y^2-3y+1=0 且 x不等于y 求x^4+y^4的值
答:
x^2-3x+1=0
y^2-3y+1=0
所以:x和y是方程a^2-3a+1=0的两个根
根据韦达定理:
x+y=3
xy=1
所以:
(x+y)^2=9
x^2+2xy+y^2=9
所以:
x^2+y^2=7
所以:
(x^2+y^2)^2=49
x^4+2(xy)^2+y^4=49
x^4+y^4+2=49
解得:
x^4+y^4=47
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