數學不等式問題......

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

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 )

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 !!

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.