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[代數][謝謝老師]
版主: thepiano
Re: [代數]
a + 3b = 74
2a + c = 103
4a + 2e = 186
b = - (a - 74) / 3
c = 103 - 2a
e = 93 - 2a
(a + b) + (3a + d) = 2(2a + c)
d = 2c - b
(2a + c) + (4a + e) = 2(3a + d)
d = (c + e) / 2
2(103 - 2a) + (a - 74) / 3 = (103 - 2a + 93 - 2a) / 2
a = 50,c = 3,b = 8,d = -2
3a + 4d = 142
2a + c = 103
4a + 2e = 186
b = - (a - 74) / 3
c = 103 - 2a
e = 93 - 2a
(a + b) + (3a + d) = 2(2a + c)
d = 2c - b
(2a + c) + (4a + e) = 2(3a + d)
d = (c + e) / 2
2(103 - 2a) + (a - 74) / 3 = (103 - 2a + 93 - 2a) / 2
a = 50,c = 3,b = 8,d = -2
3a + 4d = 142
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