高中數學
1.已知a.b.c為相異實數且a+b+c≠0,若方程式(a-b)x²+(b-c)x+(c-a)=0的二根相等,則2a+b+c/5a+b+c的值為?
2.x為實數,若i(x-i)⁴為實數,求x=?
(i=√-1)
希望能附上算式,謝謝?
回答 (2)
1.已知a,b,c為相異實數且a+b+c≠0,若方程式(a-b)x^2+(b-c)x+(c-a)=0的二根相等,則(2a+b+c)/(5a+b+c)的值為?
Sol
D=(b-c)^2-4(a-b)(c-a)=0
(c-b)^2-4(a-b)(c-a)=0
[(a-b)+(c-a)]^2-4(a-b)(c-a)=0
[(a-b)-(c-a)]^2=0
2a=b+c
(2a+b+c)/(5a+b+c)=4a/(7a)=4/7
2.x為實數,若i(x-i)^4為實數,求x=?
Sol
i(x-i)^4=a
(x-i)^4=-ai
x^4-4x^3i+6x^2i^2-4xi^3+i^4=-ai
x^4-4x^3i-6x^2+4xi+1=-ai
(x^4-6x^2+1)+i(-4x^3+4x)=-ai
x^4-6x^2+1=0
x^2=[6+/-√20]/2=[6+/-2√5]/2
=(√5+/-1)^2/2
(1) x^2=(√5-1)^2/2
x=+/-(√5-1)/√2=+/-(√10-√2)/2
(2) x^2=(√5+1)^2/2
x=+/-(√5+1)/√2=+/-(√10+√2)/2
收錄日期: 2021-04-18 17:35:22
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