对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0}...对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0},是否存在实数a,使A∪B≠空集?若a不存在,请说
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/18 07:43:13
![对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0}...对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0},是否存在实数a,使A∪B≠空集?若a不存在,请说](/uploads/image/z/4487089-49-9.jpg?t=%E5%AF%B9%E4%BA%8E%E5%AE%9E%E6%95%B0%E9%9B%86A%3D%7Bx%E4%B8%A8x%26%23178%3B-2ax%2B%EF%BC%884a-3%EF%BC%89%3D0%7D%E5%92%8CB%3D%7Bx%E4%B8%A8x%26%23178%3B-2%E2%88%9A2+ax%2B%EF%BC%88a%26%23178%3B%2Ba%2B2%EF%BC%89%3D0%7D...%E5%AF%B9%E4%BA%8E%E5%AE%9E%E6%95%B0%E9%9B%86A%3D%7Bx%E4%B8%A8x%26%23178%3B-2ax%2B%EF%BC%884a-3%EF%BC%89%3D0%7D%E5%92%8CB%3D%7Bx%E4%B8%A8x%26%23178%3B-2%E2%88%9A2+ax%2B%EF%BC%88a%26%23178%3B%2Ba%2B2%EF%BC%89%3D0%7D%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E6%95%B0a%2C%E4%BD%BFA%E2%88%AAB%E2%89%A0%E7%A9%BA%E9%9B%86%3F%E8%8B%A5a%E4%B8%8D%E5%AD%98%E5%9C%A8%2C%E8%AF%B7%E8%AF%B4)
对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0}...对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0},是否存在实数a,使A∪B≠空集?若a不存在,请说
对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0}...
对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0},是否存在实数a,使A∪B≠空集?若a不存在,请说明理由;若a存在,请求出实数a的取值范围.
对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0}...对于实数集A={x丨x²-2ax+(4a-3)=0}和B={x丨x²-2√2 ax+(a²+a+2)=0},是否存在实数a,使A∪B≠空集?若a不存在,请说
A∪B≠空集,只要A,B不同为空集就可以
通过判别式求A,B中方程有解的取值范围,然后取两者的并集.如果并集不为空集,则这样的a存在.
其实解的过程应该不难.