標題:

F1 Maths ~~~ Inequalities

發問:

1) x + 3 > 10 2) 8x < 48 3) 11 + 8x > 32 4) 2(x-1) > 3 5) Find the two smallest consecutive integers whose sum is greater than 35.

最佳解答:

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1) x + 3 > 10 x + 3 -3 > 10 - 3 x > 7 2) 8x < 48 8x/8< 48/8 x < 6 3) 11 + 8x > 32 11 + 8x - 11> 32 - 11 8x > 21 8x/8 > 21/8 x > 21/8 4) 2(x - 1) > 3 2(x - 1)/2 > 3/2 x - 1 > 3/2 x - 1 + 1 > (3/2) + 1 x > 5/2 5) Let n and (n + 1) be the two consecutive integers. n + (n + 1) > 35 2n + 1 > 35 2n + 1 - 1 > 35 - 1 2n > 34 n > 17 smallest n = 18 smallest (n + 1) = 19 The two smallest consecutive integers are 18 and 19.

其他解答:

1)x+3>10 =x>7 2)8x<48 =632 =8x>32-11 =8x>21 =x>21/8 4)2(x-1)>3 =2x-2>3 =2x>5 =x=2.5 5)18+19 =37 http://www.wolframalpha.com
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