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1/(x1的三次方)+1/(x2的三次方)!求设x1,x2是方程2(x的平方)-6x+3=0的两个根,利用根与系数的关系,求式子的值!
题目详情
1/(x1的三次方)+1/(x2的三次方)!求
设x1,x2是方程2(x的平方)-6x+3=0的两个根,利用根与系数的关系,求式子的值!
设x1,x2是方程2(x的平方)-6x+3=0的两个根,利用根与系数的关系,求式子的值!
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答案和解析
2x²-6x+3=0
x1+x2=-(-6)/2=3
x1x2=3/2
1/x1³+1/x2³
=(x1³+x³)/(x1x2)³
=(x1+x2)(x1²-x1x2+x2²)/(x1x2)³
=3((x1+x2)²-3x1x2)/(3/2)³
=(3²-3*3/2)/(9/8)
=8(1-1/2)
=4
x1+x2=-(-6)/2=3
x1x2=3/2
1/x1³+1/x2³
=(x1³+x³)/(x1x2)³
=(x1+x2)(x1²-x1x2+x2²)/(x1x2)³
=3((x1+x2)²-3x1x2)/(3/2)³
=(3²-3*3/2)/(9/8)
=8(1-1/2)
=4
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