Question

高一数学题，求解答

3题，4题，6题第一问

Answer 1

3、由韦达定理，x1+x2=3，x1x2=-1/2，所以(1-x1)(1-x2)=1-(x1+x2)+x1x2=-5/2
4、由韦达定理，-b/a=4/3，c/a=-1/3，取特殊值a=3，则b=-4，c=-1，即3x²-4x-1=0
6、|a-b|²=(a+b)²-4ab=44，所以|a-b|=2√11

Answer 2

(1.)
x1*x2=-1/2，x1+x2=3，
(1-2*x1)(1-2*x2)=1-2*x1-2*x2+4*x1*x2=1-2(x1+x2)+4*x1*x2=-7
(2.)
令a=-3，b=4，c=1
-3x^2+4x+1=0
(3.)
a+b=8，ab=5，则(a+b)^2=a^2+2ab+b^2=64

(|a-b|)^2=a^2-2ab-b^2=64-10-10=44
|a-b|=2√11

Answer 3

(3)
2x^2-6x-1=0
x1+x2=6/2=3
x1.x2=-1/2
(1-2x1)(1-2x2)
=1-2(x1+x2)+4x1.x2
=1-2(3)+4(-1/2)
=1-6-2
=-7
(4)
u,v, 为一元二次方程
u+v=4/3
uv=-1/3
3x^2-4x-1=0
(5)
x^2+√2.x+a=0
1-√2 为根
另一个根=x2
1-√2 + x2 =-√2
x2 =-1
另一个根=x2 =-1
f(x) =x^2+√2.x+a
f(1-√2) = 0
(1-√2)^2 +√2.(1-√2)+a=0
3-2√2 +√2 -2 +a =0
1-√2+a =0
a=√2-1
(6)
a≠b
a+b=8
ab=5
(1)
(a-b)^2
=(a+b)^2 -4ab
=64-20
44
|a-b|=√44 =2√11
(2)
(b-1)/(a-1)+ (a-1)/(b-1)
=[(b-1)^2+(a-1)^2]/[(a-1)(b-1)]
=( a^2+b^2-2(a+b) +2 )/( ab-(a+b) +1)
=[(a+b)^2-2ab-2(a+b) +2 )/( ab-(a+b) +1)
=(64 -10-16+2)/(5-8+1)
=40/(-2)
=-20