函数y=x/kx2+kx+1的定义域为R,则实数k的取值范围为y=x/(kx²+kx+1)①k=0时y=x/1=x,符合②k≠0时要满足kx²+kx+1≠0对任意x都成立所以k>0,Δ=k²-4k
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![函数y=x/kx2+kx+1的定义域为R,则实数k的取值范围为y=x/(kx²+kx+1)①k=0时y=x/1=x,符合②k≠0时要满足kx²+kx+1≠0对任意x都成立所以k>0,Δ=k²-4k](/uploads/image/z/1265781-21-1.jpg?t=%E5%87%BD%E6%95%B0y%3Dx%2Fkx2%2Bkx%2B1%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAR%2C%E5%88%99%E5%AE%9E%E6%95%B0k%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E4%B8%BAy%3Dx%2F%28kx%26%23178%3B%2Bkx%2B1%29%E2%91%A0k%3D0%E6%97%B6y%3Dx%2F1%3Dx%2C%E7%AC%A6%E5%90%88%E2%91%A1k%E2%89%A00%E6%97%B6%E8%A6%81%E6%BB%A1%E8%B6%B3kx%26%23178%3B%2Bkx%2B1%E2%89%A00%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E9%83%BD%E6%88%90%E7%AB%8B%E6%89%80%E4%BB%A5k%3E0%2C%CE%94%3Dk%26%23178%3B-4k)
函数y=x/kx2+kx+1的定义域为R,则实数k的取值范围为y=x/(kx²+kx+1)①k=0时y=x/1=x,符合②k≠0时要满足kx²+kx+1≠0对任意x都成立所以k>0,Δ=k²-4k
函数y=x/kx2+kx+1的定义域为R,则实数k的取值范围为
y=x/(kx²+kx+1)
①k=0时y=x/1=x,符合
②k≠0时
要满足kx²+kx+1≠0对任意x都成立
所以k>0,Δ=k²-4k
函数y=x/kx2+kx+1的定义域为R,则实数k的取值范围为y=x/(kx²+kx+1)①k=0时y=x/1=x,符合②k≠0时要满足kx²+kx+1≠0对任意x都成立所以k>0,Δ=k²-4k
有啊
2里面k>0,Δ=k²-4k<0
不就是b2-4ac<0的情况吗
b2-4ac<0就是△<0a
楼主想问什么