第二讲归纳总结-【成才之路】2020-2021学年高中新课程数学同步学习指导（人教A版选修4-5）

数学 （选修 4 - 5·人教 A 版）　 ∴ 此时 p 的取值范围是{p | p≤ - 3 或 p≥ 3 2 },取补集即得所 求实数 p 的范围,即{p | - 3 < p < 3 2 }. 5. C　 因为△A2 B2 C2 内角的正弦值均为正值,所以△A1 B1 C1 的 三个内角的余弦值均为正值,所以△A1 B1 C1 为锐角三角形. 设 sinA2 = cosA1 ,sinB2 = cosB1 ,sinC2 = cosC1 ,则 sinA2 = cosA1 = sin( π 2 - A1 ),sinB2 = cosB1 = sin( π 2 - B1 ), sinC2 = cosC1 = sin( π 2 - C1 ). 若△A2 B2 C2 为锐角三角形, 则 A2 + B2 + C2 = ( π 2 - A1 ) + ( π 2 - B1 ) + ( π 2 - C1 ) = π 2 ,与 三角形内角和为 π 矛盾. 由于若△A2 B2 C2 为直角三角形,显然不成立, 故△A2 B2 C2 为钝角三角形. 故选 C. 6. A≥ n A = 1 1 + 1 2 + 1 3 + … + 1 n ≥ 1 n + 1 n + … + 1 n ìî íï ï ï ï ïn项 = n n = n. 7. lg8·lg12 < 1 lg8·lg12 < (lg8 + lg12 2 )2 = (lg96 2 )2 < (lg100 2 )2 = 1. 8. 证明:假设 4a(1 - b)、4b(1 - c)、4c(1 - d)、4d(1 - a) 都大于 1,则 a(1 - b) > 1 4 ,b(1 - c) > 1 4 ,c(1 - d) > 1 4 ,d(1 - a) > 1 4 . ∴ a(1 - b) > 1 2 , b(1 - c) > 1 2 , c(1 - d) > 1 2 , d(1 - a) > 1 2 . 又 a(1 - b) ≤ a + (1 - b) 2 , b(1 - c) ≤ b + (1 - c) 2 , c(1 - d)≤c + (1 - d) 2 , d(1 - a)≤d + (1 - a) 2 , ∴ a + (1 - b) 2 > 1 2 , b + (1 - c) 2 > 1 2 , c + (1 - d) 2 > 1 2 , d + (1 - a) 2 > 1 2 , 以上四个式子相加,得 2 > 2,矛盾. ∴ 原命题结论成立. 9. (1)令 n = 1 得:S21 - ( - 1)S1 - 3 × 2 = 0,即 S 2 1 + S1 - 6 = 0, ∴ (S1 + 3)(S1 - 2) = 0, ∵ S1 > 0,∴ S1 = 2,即 a1 = 2. (2)由 S2n - (n 2 + n - 3)Sn - 3(n 2 + n) = 0,得:(Sn + 3) [Sn - (n2 + n)] = 0, ∵ an > 0(n∈N ∗ ),Sn > 0,从而 Sn + 3 > 0,∴ Sn = n 2 + n, ∴ 当 n≥2 时,an = Sn - Sn - 1 = n 2 + n - [(n - 1)2 + (n - 1)] = 2n, 又 a1 = 2 = 2 × 1,∴ an = 2n(n∈N ∗ ). (3)当 k∈N∗ 时,k2 + k 2 > k2 + k 2 - 3 16 = (k - 1 4 )(k + 3 4 ), ∴ 1 ak(ak + 1) = 1 2k(2k + 1) = 1 4 · 1 k(k + 1 2 ) < 1 4 · 1 (k - 1 4 )(k + 3 4 ) = 1 4 · 1 (k - 1 4 )·[(k + 1) - 1 4 ] = 1 4 ·[ 1 k - 1 4 - 1 (k + 1) - 1 4 ] ∴ 1 a1 (a1 + 1) + 1 a2 (a2 + 1) + … + 1 an(an + 1) < 1 4 [ ( 1 1 - 1 4 - 1 2 - 1 4 ) + ( 1 2 - 1 4 - 1 3 - 1 4 ) + … + ( 1 n - 1 4 - 1 (n + 1) - 1 4 ) ] = 1 4 ( 1 1 - 1 4 - 1 (n + 1) - 1 4 ) = 1 3 - 1 4n + 3 < 1 3 . 本讲归纳总结 　 　 典例试做 1:∵ a + b + c = 1, 3(a2 + b2 + c2 ) - (a + b + c)2 = 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = (a - b)2 + (b - c)2 + (c - a)2 ≥0, ∴ 3(a2 + b2