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.
(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.
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.
(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.