已知集合A={(x,y)|x²+mx-y+2=0,x∈R},B={(x,y)|x-y+1=0,0≤x≤2}.若A∩B≠空集,求m的取值范围
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![已知集合A={(x,y)|x²+mx-y+2=0,x∈R},B={(x,y)|x-y+1=0,0≤x≤2}.若A∩B≠空集,求m的取值范围](/uploads/image/z/7011942-6-2.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7B%28x%2Cy%29%7Cx%26%23178%3B%2Bmx-y%2B2%3D0%2Cx%E2%88%88R%7D%2CB%3D%7B%28x%2Cy%29%7Cx-y%2B1%3D0%2C0%E2%89%A4x%E2%89%A42%7D.%E8%8B%A5A%E2%88%A9B%E2%89%A0%E7%A9%BA%E9%9B%86%2C%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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已知集合A={(x,y)|x²+mx-y+2=0,x∈R},B={(x,y)|x-y+1=0,0≤x≤2}.若A∩B≠空集,求m的取值范围
已知集合A={(x,y)|x²+mx-y+2=0,x∈R},B={(x,y)|x-y+1=0,0≤x≤2}.若A∩B≠空集,求m的取值范围
已知集合A={(x,y)|x²+mx-y+2=0,x∈R},B={(x,y)|x-y+1=0,0≤x≤2}.若A∩B≠空集,求m的取值范围
因为A∩B≠空集,所以二式联立应当有解,
将x²+mx-y+2=0和x-y+1=0联立,消去TY,即可得到关于X的二元一次方程,
因为该方程有解,所以令判别式大于等于0即可求出m的取值范围