a22b=c2d2
1a(4)a2a4
1bb2b4
2a+12b+12c+12d+11cc2c4
1dd2d4
222222=0.22
=(a b)(a c)(a d)(b c)(b d)(c d)(a+b+c+d);证明
1aa2a4
1bb2b4
1cc2c4
1dd2d4
1110b ac ad a=0b(b a)c(c a)d(d a)0b2(b2 a2)c2(c2 a2)d2(d2 a2)
111
=(b a)(c a)(d a)bcd
222
(b+a)c(c+a)d(d+a)11
=(b a)(c a)(d a)0c bd b
0c(c b)(c+b+a)d(d b)(d+b+a)1=(b a)(c a)(d a)(c b)(d b)c(c+1b+a)d(d+b+a)=(a b)(a c)(a d)(b c)(b d)(c d)(a+b+c+d).
x
0(5)
0an 1x 0an 10 1 0an 2
0000
=xn+a1xn 1+ +an 1x+an.x 1a2x+a1