那位大大願意教我 這兩的算式?
回答 (1)
2020-10-17 1:41 pm
|x-a| ≦ b <==> -b ≦ x-a ≦ b <==> a-b ≦ x ≦ a+b
∴ a-b = -2, a+b = 8
∴ a = 3, b = 5.
|ax+3| ≦ b <==> -b ≦ ax + 3 ≦ b <==> -3-b ≦ ax ≦ -3+b
設 a > 0 則解為
(-3-b)/a ≦ x ≦ (-3+b)/a
則 -a = -3 -b, 4a = -3 + b
則 3a = -6 與 a > 0 矛盾.
a < 0,
∴ -3-b ≦ ax ≦ -3+b <==> (-3+b)/a ≦ x ≦ (-3-b)/a
∴ -3 + b = -a, -3 -b = 4a
∴ -6 = 3a
∴ a = -2
∴ b = 5
∴ a-b = -2, a+b = 8
∴ a = 3, b = 5.
|ax+3| ≦ b <==> -b ≦ ax + 3 ≦ b <==> -3-b ≦ ax ≦ -3+b
設 a > 0 則解為
(-3-b)/a ≦ x ≦ (-3+b)/a
則 -a = -3 -b, 4a = -3 + b
則 3a = -6 與 a > 0 矛盾.
a < 0,
∴ -3-b ≦ ax ≦ -3+b <==> (-3+b)/a ≦ x ≦ (-3-b)/a
∴ -3 + b = -a, -3 -b = 4a
∴ -6 = 3a
∴ a = -2
∴ b = 5
收錄日期: 2021-05-04 02:31:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201017044359AAaLOwz