1) 解二次方程,並以a+bi的形式表示答案
2x(2-x)=3
2) 求實數x和y的值
(x-1)+7i=4-(y+2)i
3) 求實數x和y的值
10+14i=2xi^3-yi^2
解F5數學題 15點**
2012-09-22 10:53 pm
回答 (2)
2012-09-23 1:44 am
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2012-09-23 14:07:56 補充：
1)
2x(2 - x) = 3
4x - 2x² = 3
2x² - 4x + 3 = 0
3)
10 + 14i = 2xi³ - yi²
10 + 14i = y - 2xi
14i = - 2xi and y = 10
14 = - 2x and y = 10
x = - 7 and y = 10
參考: My Maths World
2012-09-23 5:48 pm
(1) 2x(2-x)=3
4x - 2x^2 = 3
0 = 3 - 4x + 2x^2
+ 2x^2 - 4x + 3 = 0
(3)
10+14i = 2xi^3 - yi^2
10 + 14 i = 2xi(-1) - y(-1)
10 + 14 i = -2xi + y
10 + 14 i = + y - 2xi
10 + 14 i = + y + (-2x) i
10 = y, +14 = -2x
y = 10; 2x = -14, x = - 7
4x - 2x^2 = 3
0 = 3 - 4x + 2x^2
+ 2x^2 - 4x + 3 = 0
(3)
10+14i = 2xi^3 - yi^2
10 + 14 i = 2xi(-1) - y(-1)
10 + 14 i = -2xi + y
10 + 14 i = + y - 2xi
10 + 14 i = + y + (-2x) i
10 = y, +14 = -2x
y = 10; 2x = -14, x = - 7
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