)(x-x2)…(x-xn),m=degg,则 Res(f, g+hf (9(x)+h(x1)f(x1) =9(x)=Res(f,9 5.设f(x)=a0xn+a1xm-1+…+an-1x+an∈K[x] 证明:f(x)的判别式 (f)=(-1) e(f,∫) 证明D()=∞m-2Ⅱ(x1-x)2=(-1)an-2Ⅱ(x1-x) 1)(x1-xi+1)…( (-1)=2a-2Ⅱf"(x;) n(n=1 ao Res(, f) 6.计算下列多项式的结式: (2)f(x)=2x3-3x2-x+2,g(x)=x4-2x3-3x+4 (3)f(x)=x+x+1,g(x)=x2-3x+2; (4)∫(x)=xn+1,g(x)=(x-1) )f)=x=1,)== (6)f(x)=a0x+a1x-1+…+an-1x+an g(r) +a1xn-2+…+ 解()R(9)=(-123R(2-x-1,=(-1)0.2.(-) (2)f(x),g(x)有公共根1,所以结式Res(f,g)=0. (3)Res(f,g)=(-1)2n(1+1+1)(2+2+1)=3(2+3) (4)Res(f,g)=(-1)n.2 )()如(mn)=d>1,则二与二有公共根因此Rc(,9)=0 (b)如(m,n)=1,不妨设n>m,则n=mq+r,0≤r<m.显然(m,r)=1.则 1 rmqr'r-l(rmq-1z+ar-1 x-1 从而 Ix-1)=(-1ym-1)m+)Res 1mm-1 我们证明(m-1)(n+r)一定是偶数 如m-1是偶数,则结论成立.现设m-1是奇数,则m为偶数,从而n是奇数,r也是奇数,于是 n+r是偶数从而✭✒✮: ✎ f(x) = (x − x1)(x − x2)· · ·(x − xn), m = deg g, ✪ Res(f, g + hf) = Yn i=1 (g(xi) + h(xi)f(xi)) = Yn i=1 g(xi) = Res(f, g). 5. ✎ f(x) = a0x n + a1x n−1 + · · · + an−1x + an ∈ K[x], ✜✣✢: f(x) ✕✒➏✳✛ D(f) = (−1) n(n−1) 2 a −1 0 Res(f, f 0 ). ✭✒✮: D(f)= a 2n−2 0 Q 16i<j6n (xi − xj ) 2 = (−1) n(n−1) 2 a 2n−2 0 Q i6=j (xi − xj ) = (−1) n(n−1) 2 a 2n−2 0 Qn i=1 (xi − x1)· · ·(xi − xi−1)(xi − xi+1)· · ·(xi − xn) = (−1) n(n−1) 2 a n−2 0 Qn i=1 f 0 (xi) = (−1) n(n−1) 2 a n−2 0 Res((x − x1)· · ·(x − xn), f 0 ) = (−1) n(n−1) 2 a −1 0 Res(f, f 0 ). 6. ❂✒❃✒❄✒❅✒✙✒✚✒✛✒✕❹ ✛: (1) f(x) = x 3 − 3x 2 + 2x + 1, g(x) = 2x 2 − x − 1; (2) f(x) = 2x 3 − 3x 2 − x + 2, g(x) = x 4 − 2x 3 − 3x + 4; (3) f(x) = x n + x + 1, g(x) = x 2 − 3x + 2; (4) f(x) = x n + 1, g(x) = (x − 1)n; (5) f(x) = x n − 1 x − 1 , g(x) = x m − 1 x − 1 ; (6) f(x) = a0x n + a1x n−1 + · · · + an−1x + an, g(x) = a0x n−1 + a1x n−2 + · · · + an−2x + an−1. ❇ : (1) Res(f, g) = (−1)2·3 Res(2x 2 − x − 1, f) = (−1)6 · 2 3 · f  − 1 2  f(1) = −7. (2) f(x), g(x) q✒➐✒➑❢ 1, ❚✒❯❹ ✛ Res(f, g) = 0. (3) Res(f, g) = (−1)2n(1 + 1 + 1)(2n + 2 + 1) = 3(2n + 3). (4) Res(f, g) = (−1)n · 2 n. (5) (a) ✤ (m, n) = d > 1, ✪ x n − 1 x − 1 ★ x m − 1 x − 1 q✒➐✒➑❢, ✹✒✺ Res(f, g) = 0. (b) ✤ (m, n) = 1, ✽✒➒✒✎ n > m, ✪ n = mq + r, 0 6 r < m. ❼✒❽ (m, r) = 1. ✪ x n − 1 x − 1 = x mqx r − 1 x − 1 = (x mq − 1)x r + x r − 1 x − 1 , ✉ ✾ Res  x n − 1 x − 1 , x m − 1 x − 1  = (−1)m−1)(n+r) Res  x r − 1 x − 1 , x m − 1 x − 1  . ➓✒➔✜✣✢ (m − 1)(n + r) P✒⑤✒✏✒→✒✑. ✤ m − 1 ✏✒→✒✑, ✪❹✒➣❤✒↔. ↕✒✎ m − 1 ✏✒➙✒✑, ✪ m ✴→✒✑, ✉ ✾ n ✏✒➙✒✑, r ✫✒✏✒➙✒✑, ■✏ n + r ✏✒→✒✑. ✉ ✾ Res  x n − 1 x − 1 , x m − 1 x − 1  = Res  x r − 1 x − 1 , x m − 1 x − 1  . · 7 ·