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I don't know how to find the range of the values of x?
回答 (1)
2015-11-08 4:11 am
(1) y = x^2 - 8x + 14
2(x + y) = 16 ==> x = 8 - y
x = 8 - y >= y + 2
0 >= 2y - 6
0 >= y - 3
= x^2 - 8x + 14 - 3
= x^2 - 8x + 11
= (x-4+√5)(x-4-√5)
Answer: 4-√5 <= x <= 4+√5
(2) x = ?
ΔABC = 0.5(x+3)(2x-1)*sin30
= (2x^2 + 5x - 3)/4
<= 7
0 >= 2x^2 + 5x - 31
= [x+(5+√273)/4][x+(5-√273)/4]
Answer: (-5-√273)/4 <= x <= (-5+√273)/4
2(x + y) = 16 ==> x = 8 - y
x = 8 - y >= y + 2
0 >= 2y - 6
0 >= y - 3
= x^2 - 8x + 14 - 3
= x^2 - 8x + 11
= (x-4+√5)(x-4-√5)
Answer: 4-√5 <= x <= 4+√5
(2) x = ?
ΔABC = 0.5(x+3)(2x-1)*sin30
= (2x^2 + 5x - 3)/4
<= 7
0 >= 2x^2 + 5x - 31
= [x+(5+√273)/4][x+(5-√273)/4]
Answer: (-5-√273)/4 <= x <= (-5+√273)/4
收錄日期: 2021-04-18 14:10:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151107041344AA2IKIR