若a,b是方程x^2-3x-2=0的根,試求下列各式的值。
(a) a-b
(b) a^2 - b^2
(c) a^4 -b^4
a-maths
2007-06-18 4:12 am
回答 (2)
2007-06-18 4:47 am
✔ 最佳答案
若a,b是方程x2-3x-2=0的根,試求下列各式的值。則兩根和
a + b = 3
兩根積
ab = -2
(a) a-b
= √(a – b)2 = √[a2 – 2ab + b2] = √[a2 + 2ab + b2 – 4ab] = √[(a+b)2 – 4ab]
=√[(3)2 – 4(-2)]
=√17
(b) a2 - b2
= (a+b)(a-b)
= 3√17
(c) a4 -b4
= (a2 + b2)(a2 – b2)
= [(a+b)2 -2ab](a2 – b2)
= [(3)2 -2(-2)]( 3√17)
= 39√17
2007-06-19 5:49 pm
若a,b是方程x^2-3x-2=0的根
a) a+b=-(-3)=3 - - - 1
ab=-2 - - -2
(1)^2-4(2)
(a-b)^2=9-(-8)=17
a-b=正負√17
b) a^2 - b^2
=(a+b)(a-b)
=3x正負√17
(1)^2-2(2)
a^2+b^2=9-(-4)=13
c) a^4 -b^4
=(a^2+b^2)(a^2-b^2)
=13(3x正負√17)
=39正負√17
a) a+b=-(-3)=3 - - - 1
ab=-2 - - -2
(1)^2-4(2)
(a-b)^2=9-(-8)=17
a-b=正負√17
b) a^2 - b^2
=(a+b)(a-b)
=3x正負√17
(1)^2-2(2)
a^2+b^2=9-(-4)=13
c) a^4 -b^4
=(a^2+b^2)(a^2-b^2)
=13(3x正負√17)
=39正負√17
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