问题
解答题
x2-13x+1=0,试确定x4+x-4的个位数.
答案
∵x2-13x+1=0,
∴x+
=13,1 x
∴x2+
=(x+1 x2
)2-2,1 x
=132-2,
=167,
∴x4+
=(x2+1 x4
)2-2,1 x2
=1672-2,
=278897,
∴x4+x-4的个数字是7.
答:x4+x-4的个位数为7.
x2-13x+1=0,试确定x4+x-4的个位数.
∵x2-13x+1=0,
∴x+
=13,1 x
∴x2+
=(x+1 x2
)2-2,1 x
=132-2,
=167,
∴x4+
=(x2+1 x4
)2-2,1 x2
=1672-2,
=278897,
∴x4+x-4的个数字是7.
答:x4+x-4的个位数为7.