3.3.1 二元一次不等式(组)与平面区域（练案）-【成才之路】2020-2021学年高中新课程数学同步学习指导（人教A版必修5）

▲ 199 ▲ ▲ 200 ▲ 10. (an + bn ) - (an - 1 b + abn - 1 ) = an - 1 (a - b) + bn - 1 (b - a) = (a - b)(an - 1 - bn - 1 ), (1)当 a >b >0 时,an -1 >bn -1,∴ (a -b)(an -1 -bn -1) >0, (2)当0 <a <b 时,an -1 <bn -1,∴ (a -b)(an -1 -bn -1) >0, ∴ 对任意 a > 0,b > 0,a≠b, 总有(a - b)(an - 1 - bn - 1 ) > 0. ∴ an + bn > an - 1 b + abn - 1 . B 级　 素养提升 1. D　 解法一:由 a + | b | < 0 知,a < 0,0≤ | b | < - a, ∴ b2 < a2 ,∴ a2 - b2 > 0; ∵ | b | ≥b,∴ a + b≤a + | b | < 0; ∵ | b | ≥ - b,∴ a - b≤a + | b | < 0; ∵ - a > | b | ≥b,∴ ( - a)3 > b3 ,∴ a3 + b3 < 0. ∴ A、B、C 错,D 正确. 解法二:取 a = - 2,b = ± 1,易知 a - b < 0,a3 + b3 < 0,a2 - b2 > 0,排除 A、B、C,故选 D. 2. C　 解法一:由 a > b > 0⇒0 < 1 a < 1 b ⇒a + 1 b > b + 1 a , 故选 C. 解法二:(特值法)令 a = 2,b = 1,排除 A、D,再令 a = 1 2 , b = 1 3 ,排除 B. 3. B　 ∵ f(x) = x3 是单调递增函数,x1 < - x2 ,x2 < - x3 ,x3 < - x1 , ∴ f ( x1 ) < f ( - x2 ), f ( x2 ) < f ( - x3 ), f(x3 ) < f( - x1 ), 又∵ f(x)为奇函数, ∴ f(x1 ) < - f(x2 ),f(x2 ) < - f(x3 ),f(x3 ) < - f(x1 ), ∴ f(x1) + f(x2) <0,f(x2) + f(x3) <0,f(x3) + f(x1) <0 ∴ f(x1 ) + f(x2 ) + f(x3 ) < 0. 4. B　 ∵ x > 0,y > 0,∴ x + y + 1 > 1 + x > 0,1 + x + y > 1 + y > 0, ∴ x 1 + x + y < x 1 + x , y 1 + x + y < y 1 + y , 故 M = x + y 1 + x + y = x 1 + x + y + y 1 + x + y < x 1 + x + y 1 + y = N, 即 M < N. 故选 B. 5. (2,1, - 1, - 2)　 由 a b > c d > 0 知,a、b 同号,c、d 同号, 且 a b - c d = ad - bc bd > 0. 由 ad < bc,得 ad - bc < 0,所以 bd < 0. 所以在取(a、b、c、d)时只需满足以下条件即可: ①a、b 同号,c、d 同号,b、d 异号;②ad < bc. 令 a > 0,b > 0,c < 0,d < 0,不妨取 a = 2,b = 1,c = - 1, 则 d < bc a = - 1 2 ,取 d = - 2,则(2,1, - 1, - 2)满足要求. 6. p < r < s < q　 解法一:取 a = 4,b = 2,m = 3,n = 1,则 p = 1 2 , q = 2,r = 5 7 ,s = 5 3 则 p < r < s < q(特值探路). 解法二:p - r = b a - b + m a + m = (b - a)m a(a + m) < 0,∴ p < r. ∵ a > b > 0,m > 0,n > 0 ∴ a + m > b + m > 0. a + n > b + n > 0, ∴ b + m a + m < 1,a + n b + n > 1,∴ r < s. 或 r - s = b + m a + m - a + n b + n = (b - a)(b + a + m + n) (a + m)(b + n) < 0. ∴ r < s. s - q = a + n b + n - a b = (b - a)·n b(b + n) < 0, ∴ s < q. ∴ p < r < s < q. 7. (1)∵ c > a > b > 0∴ c - a > 0,c - b > 0, 由 a > b >