標題:
續二次方程(複數)(27-30題)
發問:
最佳解答:
27a) 5-i/(1+i)(2+3i) = 5-i /(2+ 3i + 2i - 3) * i x i = -1* = 5-i /(5i-1) =(5-1)(5i+1) / (-25-1) =(10+24i) / (-26) =-5/13 - 12i/13 b) (3+i)(2-i)/4+3i = (6 -3i + 2i +1) / (4+3i) = (7- i) / (4+3i) = (7-i)(4-3i) / (16-9) =(28- 21i -4i -3)/ 7 =25/7 - 25i/7 28a)(√3 +i)(√3 -i) = 9-(-1) =10 b)(√3 + i)^3 (√3 -i)^2 =10 x 10 x (3^(1/2)+i) =100(3)^(1/2) + 100i 29a)2i- 4/3-i = (2i-4)(3+i) / (9+1) = (6i -2 -12 -4i) / 10 = (2i-14)/10 = -7/5 + i/5 29b)(2+5i)^2 + (8-10i) = 4 + 20i -25 + 8-10i =10i-13 30a)2i(-4+3i)+(4-5i)(4+5i) = -8i -6 +16 +25 =-8i-3 30b. (4-2i)^2 - (1+i)^2 =16 +16i +4 -1 -2i +1 =14i+12
其他解答:
27a) (5-i)/[(1+i)(2+3i)] = (5 - i)/(-1 + 5i) = (5 - i)(-1 - 5i)/26 = -(5/13) - (12/13)i 27b) (3+i)(2-i)/(4+3i) = (7 - i)/(4 + 3i) = (7 - i)(4 - 3i)/25 = 1 - i 28a) (根3+i)(根3-i) = 10 28b) (根3+i)^3 (根3-i)^2 = 10^2 (根3 + i) = 100根3 + 100i 29a) (2i-4)/(3-i) = 2(-2 + i)(3 + i)/10 = (-7 + i)/5 = -(7/5) + (1/5)i 29b) (2+5i)^2 + (8-10i) = 4 + 20i - 25 + 8 - 10i = -13 + 10i 30a) 2i(-4+3i) + (4-5i)(4+5i) = -6 - 8i + 16 + 25 = 35 - 8i 30b) (4-2i)^2 - (1+i)^2 = (16 - 16i - 4) - (1 + 2i - 1) = 12 - 18i
續二次方程(複數)(27-30題)
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化簡下列各數式,並以a+bi的形式表示答案27(a) 5-i/(1+i)(2+3i)(b) (3+i)(2-i)/4+3i化簡下列各數式,並以a+bi的形式表示答案28(a)化簡 (√3 +i)(√3 -i)(b)由此,化簡(√3 + i)^3 (√3 -i)^2(答案以a+bi的形式表示)化簡下列各數式,並以a+bi的形式表示答案29(a)2i- 4/3-i29b. (2+5i)^2 + (8-10i)化簡下列各數式,並以a+bi的形式表示答案30a.... 顯示更多 化簡下列各數式,並以a+bi的形式表示答案 27(a) 5-i/(1+i)(2+3i) (b) (3+i)(2-i)/4+3i 化簡下列各數式,並以a+bi的形式表示答案 28(a)化簡 (√3 +i)(√3 -i) (b)由此,化簡(√3 + i)^3 (√3 -i)^2 (答案以a+bi的形式表示) 化簡下列各數式,並以a+bi的形式表示答案 29(a)2i- 4/3-i 29b. (2+5i)^2 + (8-10i) 化簡下列各數式,並以a+bi的形式表示答案 30a. 2i(-4+3i)+(4-5i)(4+5i) 30b. (4-2i)^2 - (1+i)^2最佳解答:
27a) 5-i/(1+i)(2+3i) = 5-i /(2+ 3i + 2i - 3) * i x i = -1* = 5-i /(5i-1) =(5-1)(5i+1) / (-25-1) =(10+24i) / (-26) =-5/13 - 12i/13 b) (3+i)(2-i)/4+3i = (6 -3i + 2i +1) / (4+3i) = (7- i) / (4+3i) = (7-i)(4-3i) / (16-9) =(28- 21i -4i -3)/ 7 =25/7 - 25i/7 28a)(√3 +i)(√3 -i) = 9-(-1) =10 b)(√3 + i)^3 (√3 -i)^2 =10 x 10 x (3^(1/2)+i) =100(3)^(1/2) + 100i 29a)2i- 4/3-i = (2i-4)(3+i) / (9+1) = (6i -2 -12 -4i) / 10 = (2i-14)/10 = -7/5 + i/5 29b)(2+5i)^2 + (8-10i) = 4 + 20i -25 + 8-10i =10i-13 30a)2i(-4+3i)+(4-5i)(4+5i) = -8i -6 +16 +25 =-8i-3 30b. (4-2i)^2 - (1+i)^2 =16 +16i +4 -1 -2i +1 =14i+12
其他解答:
27a) (5-i)/[(1+i)(2+3i)] = (5 - i)/(-1 + 5i) = (5 - i)(-1 - 5i)/26 = -(5/13) - (12/13)i 27b) (3+i)(2-i)/(4+3i) = (7 - i)/(4 + 3i) = (7 - i)(4 - 3i)/25 = 1 - i 28a) (根3+i)(根3-i) = 10 28b) (根3+i)^3 (根3-i)^2 = 10^2 (根3 + i) = 100根3 + 100i 29a) (2i-4)/(3-i) = 2(-2 + i)(3 + i)/10 = (-7 + i)/5 = -(7/5) + (1/5)i 29b) (2+5i)^2 + (8-10i) = 4 + 20i - 25 + 8 - 10i = -13 + 10i 30a) 2i(-4+3i) + (4-5i)(4+5i) = -6 - 8i + 16 + 25 = 35 - 8i 30b) (4-2i)^2 - (1+i)^2 = (16 - 16i - 4) - (1 + 2i - 1) = 12 - 18i
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