1.已知A={X| 2x^2-ax+b=0},B={X| bx^2+(a+2)x+5+b=0},且A交B={1\2},求A并B2.已知集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)=0}(1)若A交B={2},求实数a的值;(2)若A并B=A,求实数a的取值范围
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![1.已知A={X| 2x^2-ax+b=0},B={X| bx^2+(a+2)x+5+b=0},且A交B={1\2},求A并B2.已知集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)=0}(1)若A交B={2},求实数a的值;(2)若A并B=A,求实数a的取值范围](/uploads/image/z/8096137-25-7.jpg?t=1.%E5%B7%B2%E7%9F%A5A%3D%7BX%7C+2x%5E2-ax%2Bb%3D0%7D%2CB%3D%7BX%7C+bx%5E2%2B%28a%2B2%29x%2B5%2Bb%3D0%7D%2C%E4%B8%94A%E4%BA%A4B%3D%7B1%5C2%7D%2C%E6%B1%82A%E5%B9%B6B2.%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7Bx%7Cx%5E2-3x%2B2%3D0%7D%2CB%3D%7Bx%7Cx%5E2%2B2%28a%2B1%29x%2B%28a%5E2-5%29%3D0%7D%281%29%E8%8B%A5A%E4%BA%A4B%3D%7B2%7D%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5A%E5%B9%B6B%3DA%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
1.已知A={X| 2x^2-ax+b=0},B={X| bx^2+(a+2)x+5+b=0},且A交B={1\2},求A并B2.已知集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)=0}(1)若A交B={2},求实数a的值;(2)若A并B=A,求实数a的取值范围
1.已知A={X| 2x^2-ax+b=0},B={X| bx^2+(a+2)x+5+b=0},且A交B={1\2},求A并B
2.已知集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)=0}
(1)若A交B={2},求实数a的值;
(2)若A并B=A,求实数a的取值范围
1.已知A={X| 2x^2-ax+b=0},B={X| bx^2+(a+2)x+5+b=0},且A交B={1\2},求A并B2.已知集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)=0}(1)若A交B={2},求实数a的值;(2)若A并B=A,求实数a的取值范围
1.因为A交B=1/2,所以1/2是两个方程的公共根
代入方程得:1/2-1/2a+b=1-a+2b=0; 1/2a+5/4b+6=2a+5b+24=0
所以a=-43/9 b=-26/9
所以集合A是18x²+43x-26=0,集合B是-26x²-25x+19=0,
用韦达定理,集合A中:x1+1/2=-43/18,1/2x1=-13/9===>x1=-26/9,x2=1/2
同理:集合B是x1=-19/13,x2=1/2
所以A并B={x|x=1/2,-26/9,-19/13}
2.A={1,2}
(1).1A∩B={2} 把x=2代入B===>2²+2(a+1)*2 +(a²-5)=0 解得a=-3或-1
再把a=-3或-1代入x²+2(a+1)x+(a²-5)=0检验,均可,所以 a=-3或-1
(2).①B={1} 1+2(a+1)+(a^2-5)=0 a=-1±根号3 检验△>0 不合,舍去
②B={2} 同1,检验,a=-1时,x=±2(舍去) 所以 a=-3
③B={1,2} 无解
所以 a=-3