题目

若x1,x2为一元二次方程x^2-5x-7=0的两个根,求x1^3+x2^3 x1^3-x2^3 x1^4=x2^4
最后一题是x1^4-x2^4 我还有别的作业，

提问时间：2021-01-19

答案

x1+x2=5,x1*x2=-7
x1-x2
=±√(x1-x2)²
=±√[(x1+x2)²-4x1*x2]
=±√(5²+4×7)
=±√53
x1³+x2³
=(x1+x2)(x1²-x1*x2+x2²)
=(x1+x2)[(x1+x2)²-3x1*x2]
=5×(5²+3×7)
=230
x1³-x2³
=(x1-x2)(x1²+x1*x2+x2²)
=(x1-x2)[(x1+x2)²-x1*x2]
=±√53×(5²+7)
=±32√53
x1^4-x2^4
=(x1²-x2²)(x1²+x2²)
=(x1-x2)(x1+x2)(x1²+x2²)
=(x1-x2)(x1+x2)[(x1+x2)²-2x1*x2]
=±√53×5×(5²+2×7)
=±195√53
方法和过程就是这样,你再检查一下,打太多字,怕打错了