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A multiplied by 5 / 6 equals B multiplied by 3 / 2 equals C multiplied by 4 / 5 equals D multiplied by 3 / 8. Please arrange the four numbers ABCD in descending order A multiplied by 5 / 6 equals B multiplied by 3 / 2 equals C multiplied by 4 / 5 equals D multiplied by 3 / 8 (ABCD is not zero). Please arrange the four numbers ABCD in descending order
5/6 a=3/2 b=4/5 c=3/8 d
It can be concluded that
b=5/9 a=40/72 a
c=25/24 a=75/72 a
d=20/9 a=160/72 a
a=72/72 a
So, when a = 0, a = b = C = D
When a > O, d > C > a > B
When AA > C > d
Arrange 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 on the circle in any order, and prove that there must be three adjacent numbers whose sum is greater than or equal to 17
Solution: first of all, the arrangement of this problem is on the circle, then, we should consider how many combinations there are. If there are 8 combinations in a row, there should be 10 combinations on the circle (the fifth grade of Jiangsu Education Press, other versions don't know what grade of knowledge it is). Secondly, we should prove that we can't give examples. Examples are a way to prove no, to prove
A number x plus 3, and then multiply by 3, is equal to the sum of 4 times of this number plus 5. What's this number? (only equations are listed)
Let this number be X. according to the meaning of the question, we can get the equation: (x + 3) × 3 = 4x + 5 & nbsp; & nbsp; & nbsp; 3x + 9 = 4x + 5 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 4
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