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標題:
數學知識交流---解方程
發問:
(1) 解方程組 a + b + ab = 1 a^2 - b^2 = -1 b>0 (2) 解方程組 a + b + ab = 1 a^2 - b^2 = -1 b<0
a + b + ab =1 ==>a = (1-b)/(1+b) Put it into a^2 - b^2 =-1, (1-b)^2/(1+b)^2 - b^2 =-1 (b^2-2b+1)-b^2(b^2+2b+1)= - b^2-2b-1 b^2 - 2b + 1 -b^4 - 2b^3 - b^2 = - b^2 - 2b -1 b^2 - b^4 - 2b^3 + 2 =0 b^4 - b^2 + 2b^3 -2=0 b^2(b^2-1)+2(b^3-1)=0 b^2(b-1)(b+1)+2(b-1)(b^2+b+1)=0 (b-1)(b^3+b^2+2b^2 + 2b + 2)=0 (b-1)(b^3+3b^2+2b + 2)=0For b>0, b^3+3b^2+2b + 2>0, so b=1, a=0 The solution is a=0, b=1For b<0, b^3+3b^2+2b + 2=0 Let f(b)=b^3+3b^2+2b + 2 Let b=x+h, f(b)=0 becomes (x+h)^3 + 2(x+h)^2 = 2(x+h)+2=0 x^3 + (3h+3)x^2 +(3h^2 +6h +2)x + (h^3+3h^2+2h+2)=0 To eliminate the x^2 term, we let h=-1 Hence, the equation becomes x^3 - x + 2 = 0 Knowing the identity x^3 + p^3 + q^3 - 3pqx= (x + p + q)(x + pw + qw^2)(x + pw^2 + qw), where w is a complex cube root of 1, we compare x^3 - x + 2 = 0 with x^3 + p^3 + q^3 - 3pqx. So, pq = 1/3 ===> q = 1/3q p^3 + q^3 = 2p^3 + 1/27q^3 = 2 p^6 -2p^3 + 1/27 = 0 p^3 = [2 ± √(2^2 - 4/27)]/2 = 1± √(26/27) We may let p = 3√[1+ √(26/27)] q = 3√[1-√(26/27)] Now, x^3 - x + 2 = 0 becomes (x + p + q)(x + pw + qw^2)(x + pw^2 + qw)=0 which has only one real root x = -3√[1+ √(26/27)]-3√[1-√(26/27)] b = x + h = x - 1 = -1 - 3√[1+ √(26/27)] - 3√[1-√(26/27)] a = (1-b)/(1+b) = -{2+ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}/{ 3√[1+ √(26/27)] + 3√[1-√(26/27)]} The solution is a = -{2+ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}/{ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}, b = -1 - 3√[1+ √(26/27)] - 3√[1-√(26/27)]
其他解答:
(a+b+ab)^2=1a^2+2ab+b^2+2ba^2+2ab^2+a^2b^2=1(a+b)^2+ab(a+b+ab)=1(a+b)^2+ab=11-ab=(a+b)^2Compare (1),1-ab=a+b a+b=(a+b)^2a+b=1So,ab=0 If b>0a=0, b=1 If b<0We have a = 0, b=?No solution.
數學知識交流---解方程
發問:
(1) 解方程組 a + b + ab = 1 a^2 - b^2 = -1 b>0 (2) 解方程組 a + b + ab = 1 a^2 - b^2 = -1 b<0
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最佳解答:a + b + ab =1 ==>a = (1-b)/(1+b) Put it into a^2 - b^2 =-1, (1-b)^2/(1+b)^2 - b^2 =-1 (b^2-2b+1)-b^2(b^2+2b+1)= - b^2-2b-1 b^2 - 2b + 1 -b^4 - 2b^3 - b^2 = - b^2 - 2b -1 b^2 - b^4 - 2b^3 + 2 =0 b^4 - b^2 + 2b^3 -2=0 b^2(b^2-1)+2(b^3-1)=0 b^2(b-1)(b+1)+2(b-1)(b^2+b+1)=0 (b-1)(b^3+b^2+2b^2 + 2b + 2)=0 (b-1)(b^3+3b^2+2b + 2)=0For b>0, b^3+3b^2+2b + 2>0, so b=1, a=0 The solution is a=0, b=1For b<0, b^3+3b^2+2b + 2=0 Let f(b)=b^3+3b^2+2b + 2 Let b=x+h, f(b)=0 becomes (x+h)^3 + 2(x+h)^2 = 2(x+h)+2=0 x^3 + (3h+3)x^2 +(3h^2 +6h +2)x + (h^3+3h^2+2h+2)=0 To eliminate the x^2 term, we let h=-1 Hence, the equation becomes x^3 - x + 2 = 0 Knowing the identity x^3 + p^3 + q^3 - 3pqx= (x + p + q)(x + pw + qw^2)(x + pw^2 + qw), where w is a complex cube root of 1, we compare x^3 - x + 2 = 0 with x^3 + p^3 + q^3 - 3pqx. So, pq = 1/3 ===> q = 1/3q p^3 + q^3 = 2p^3 + 1/27q^3 = 2 p^6 -2p^3 + 1/27 = 0 p^3 = [2 ± √(2^2 - 4/27)]/2 = 1± √(26/27) We may let p = 3√[1+ √(26/27)] q = 3√[1-√(26/27)] Now, x^3 - x + 2 = 0 becomes (x + p + q)(x + pw + qw^2)(x + pw^2 + qw)=0 which has only one real root x = -3√[1+ √(26/27)]-3√[1-√(26/27)] b = x + h = x - 1 = -1 - 3√[1+ √(26/27)] - 3√[1-√(26/27)] a = (1-b)/(1+b) = -{2+ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}/{ 3√[1+ √(26/27)] + 3√[1-√(26/27)]} The solution is a = -{2+ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}/{ 3√[1+ √(26/27)] + 3√[1-√(26/27)]}, b = -1 - 3√[1+ √(26/27)] - 3√[1-√(26/27)]
其他解答:
(a+b+ab)^2=1a^2+2ab+b^2+2ba^2+2ab^2+a^2b^2=1(a+b)^2+ab(a+b+ab)=1(a+b)^2+ab=11-ab=(a+b)^2Compare (1),1-ab=a+b a+b=(a+b)^2a+b=1So,ab=0 If b>0a=0, b=1 If b<0We have a = 0, b=?No solution.
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