问题 解答题

已知:a+b=5,ab=3.利用anbn=(ab)n、(a+b)2=a2+2ab+b2

(1)求:a2+b2的值;

(2)求:(a+1)(b+1)(a-1)(b-1)的值.

答案

(1)∵a2+b2=(a+b)2-2ab,

∴原式=52-2×3

=25-6

=19;

(2)原式=(a+1)(a-1)(b+1)(b-1)

=(a2-1)(b2-1)

=a2b2-a2-b2+1

=(ab)2-(a2+b2)+1,

则原式=9-19+1

=-9.

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