2.2 综合法与分析法-【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）　 又 x > y > 0,∴ bx > ay, ∴ bx - ay (x + a)(y + b) > 0, 即 x x + a > y y + b . B 级　 素养提升 1. B　 因为 a2 + b2 ≥2ab 且 a2 + c2 ≥2ac,b2 + c2 ≥2bc,将三式相加 得,2(a2 + b2 + c2 )≥2ab + 2bc + 2ac,即 a2 + b2 + c2 ≥1, 又因为(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac, 所以(a + b + c)2 ≥1 + 2 × 1 = 3,故选 B. 2. A　 P - Q = loga (a 3 + 1) - loga (a 2 + 1) = loga a3 + 1 a2 + 1 ,当 0 < a < 1 时,0 < a3 + 1 < a2 + 1,0 < a 3 + 1 a2 + 1 < 1, ∴ loga a3 + 1 a2 + 1 > 0,即 P - Q > 0, ∴ P > Q. 当 a > 1 时,a3 + 1 > a2 + 1 > 0,a 3 + 1 a2 + 1 > 1,∴ loga a3 + 1 a2 + 1 > 0,即 P - Q > 0, ∴ P > Q. 3. A　 1 + qm + n - ( qm + qn ) = 1 + qm + n - qm - qn = (1 - qm ) + qn(qm - 1) = (1 - qm)(1 - qn). 若 0 < q < 1,由 m,n∈N + ,知 0 < q m < 1,0 < qn < 1,∴ 1 - qm > 0,1 - qn > 0,∴ (1 - qm)(1 - qn) > 0. 若 q > 1,由 m,n∈N + ,知 q m > 1,qn > 1,∴ 1 - qm < 0,1 - qn < 0,∴ (1 - qm)(1 - qn) > 0. 综上可知 1 + qm + n - (qm + qn) > 0, 即 1 + qm + n > qm + qn. 4. A≥B　 A - B = b + a 2ab - 2 a + b = (a + b) 2 - 4ab 2ab(a + b) = (a - b) 2 2ab(a + b) . ∵ a > 0,b > 0,∴ 2ab > 0,a + b > 0, 又(a - b)2 ≥0,∴ A≥B. 5. > 　 (x + 1)(x2 - x + 1) - (x - 1)(x2 + x + 1) = x3 + 1 - (x3 - 1) = 2 > 0. 6. ≤　 2a 1 + a2 - 1 = 2a - 1 - a 2 1 + a2 = - (1 - a) 2 1 + a2 ≤0. 7. 证明:因为 a > 0,b > 0,且 a≠b,所以 a2 b + ab2 > 2ab ab,a3 + b3 > 2ab ab. 所以 a2 b + ab2 - 2ab ab > 0, a3 + b3 - 2ab ab > 0. 所以 | a2 b + ab2 - 2ab ab | - | a3 + b3 - 2ab ab | = a2 b + ab2 - 2ab ab - a3 - b3 + 2ab ab = a2 b + ab2 - a3 - b3 = a2 (b - a) + b2 (a - b) = (a - b) (b2 - a2 ) = - (a - b)2 (a + b) < 0 所以 | a2 b + ab2 - 2ab ab | < | a3 + b3 - 2ab ab | ,所以 a2 b + ab2 比 a3 + b3 接近 2ab ab. 8. 解:(1)由 f(x),g(x)的奇偶性及 f(x) + g(x) = ex, ① 得 - f(x) + g(x) = e - x. ② 联立①②解得 f(x) = 1 2 (ex - e - x), g(x) = 1 2 (ex + e - x). 当 x > 0 时,ex > 1,0 < e - x < 1,故 f(x) > 0. ③ 又由基本不等式,有 g(x) = 1 2 (ex + e - x) > exe - x = 1, 即 g(x) > 1. ④ (2)由(1)得 f′(x) = 1 2 (ex - 1 ex )′ = 1 2 (ex + e x e2x ) = 1 2 (ex + e - x) = g(x), ⑤ g′(x) = 1 2 (ex + 1 ex )′ = 1 2 (ex - e x e2x ) = 1 2 (ex - e - x) = f(x), ⑥ 当 x > 0 时,