1.若兩個連續整數的和大於45,求當中較大的整數的最少可能值。
2.若兩個連續整數的和小於19,求當中較小的整數的最大可能值。
最好就解釋埋點做......
thx~
數學不等式問題......
2010-12-05 11:05 pm
回答 (3)
2010-12-06 4:11 am
✔ 最佳答案
1. 設連續數x , x+1x + x + 1 > 45
2x > 44
x > 22
最小可能值是23
2. 設連續數x , x+1
x + x + 1 < 19
2x < 18
x < 9
最大可能值是8
1.[p^0(-q)^-3]^-2
= ( 1 ( -q^3 ) ) ^ -2
= ( -q^3 ) ^ -2
= 1 / q^6
2.(8x^2/y)^-1(x^3/4y^-1)^-2
= ( y / 8x^2 ) ( 4yx^3 ) ^ -2
= ( y / 8x^2 ) ( 1 / 16x^6y^2 )
= ( 1 / 128yx^8 )
參考: Hope I Can Help You ^_^
2010-12-05 11:23 pm
1. 設連續數x ,x+1
x+x+1 > 45
2x > 44
x > 22
最小可能值是23
2. 設連續數x ,x+1
x+x+1 < 19
2x < 18
x < 9
最大可能值是8
I hope can help you !!
x+x+1 > 45
2x > 44
x > 22
最小可能值是23
2. 設連續數x ,x+1
x+x+1 < 19
2x < 18
x < 9
最大可能值是8
I hope can help you !!
2010-12-05 11:20 pm
1) Let the two integers be x and ( x - 1 ).
x + ( x - 1 ) > 45
2x > 46
x > 23
Min (x) = 24
Hence, the minimum possible value of the larger integer is 24.
2) Let the two integers be x and ( x + 1 ).
x + ( x + 1 ) > 19
2x < 18
x < 9
Max (x) = 8
Hence, the maximum possible value of the smaller integer is 8.
x + ( x - 1 ) > 45
2x > 46
x > 23
Min (x) = 24
Hence, the minimum possible value of the larger integer is 24.
2) Let the two integers be x and ( x + 1 ).
x + ( x + 1 ) > 19
2x < 18
x < 9
Max (x) = 8
Hence, the maximum possible value of the smaller integer is 8.
參考: Mathematics Teacher Mr. Ip
收錄日期: 2021-04-13 17:40:58
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