若a和b是二次方程x^2-2x-6=0的根,求下列各式的值
a)(1/a)+(1/b)
b)a^2b+ab^2
c)(a-2)(b-2)
d)a^2+2b
我唔識d啊~help
f.4 maths
2010-10-11 4:54 am
回答 (2)
2010-10-11 5:04 am
✔ 最佳答案
a + b = - (-2) / 1 = 2ab = -6/1 = - 6a)1/a + 1/b= (a + b) / (ab)= 2 / -6= - 1/3b)a²b + ab²= ab(a + b)= (- 6)(2)= - 12c)(a - 2)(b - 2)= ab - 2b - 2a + 4= ab - 2(a + b) + 4= (- 6) - 2(2) + 4= - 6d)a² + 2b= a² + 2(a+b - a)= a² + 2(2 - a)= a² - 2a + 4= a² - 2a - 6 + 10= 0 + 10= 10
2010-10-11 5:10 am
d.
x^2-2x-6=0
a+b=2
ab=-6
a=2-b
(2-b)b=-6
-b^2+2b=-6
(a+b)^2=2^2
a^2+2ab+b^2=4
a^2+b^2-12=4
a^2+b^2=16
b^2=16-a^2
---> -(16-a^2)+2b=-6
a^2-16+2b=-6
Therefore, a^2+2b=10
x^2-2x-6=0
a+b=2
ab=-6
a=2-b
(2-b)b=-6
-b^2+2b=-6
(a+b)^2=2^2
a^2+2ab+b^2=4
a^2+b^2-12=4
a^2+b^2=16
b^2=16-a^2
---> -(16-a^2)+2b=-6
a^2-16+2b=-6
Therefore, a^2+2b=10
參考: me
收錄日期: 2021-04-21 22:17:20
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