解答4條數學方程!THX

1)
(2n - 3)/4 + (1 - 4n)/5 + 7/10 = 0
5(2n - 3)/20 + 4(1 - 4n)/20 + 14/20 = 0
(10n - 15)/20 + (4 - 16n)/20 + 14/20 = 0
(10n - 15 + 4 - 16n + 14)/20 = 0
-6n + 3 = 0
n = 1/2


= = = = =
2)
(r + 2)/8 - 3(r - 2)/6 = (10 - r)/4
(r + 2)/8 - (r - 2)/2 = (10 - r)/4
(r + 2)/8 - 4(r - 2)/8 - 2(10 - r)/8 = 0
(r + 2)/8 - (4r - 8)/8 - (20 - 2r)/8 = 0
(r + 2 - 4r + 8 - 20 + 2r)/8 = 0
-r - 10 = 0
r = -10


= = = = =
3)
Let n be the largest odd number.
Then the other two odd numbers are (n - 2) and (n - 4).

n + (n - 2) + (n - 4) = 81
3n - 6 = 81
3n = 87
n = 29

Hence, the largest odd number = 29


= = = = =
4)
Let y cm be the length of the original equilateral triangle.
Then, the length of each small equilateral triangle = y/3 cm

Total perimeter of the 9 small triangle = 9 * 3 * (y/3) cm = 9y cm
Perimeter of the original triangle = 3y cm

9y - 3y = 54
6y = 54
y = 9

Hence, the length of a side of the original triangle = 9 cm