设集合A={x²+3k²≥2k(2x-1},B={x|x²-(2x-1)k+k²≥0}且A∈B,试求k的取
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![设集合A={x²+3k²≥2k(2x-1},B={x|x²-(2x-1)k+k²≥0}且A∈B,试求k的取](/uploads/image/z/3813046-70-6.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88A%3D%7Bx%26%23178%3B%2B3k%26%23178%3B%E2%89%A52k%282x-1%7D%2CB%3D%7Bx%7Cx%26%23178%3B-%282x-1%29k%2Bk%26%23178%3B%E2%89%A50%7D%E4%B8%94A%E2%88%88B%2C%E8%AF%95%E6%B1%82k%E7%9A%84%E5%8F%96)
设集合A={x²+3k²≥2k(2x-1},B={x|x²-(2x-1)k+k²≥0}且A∈B,试求k的取
设集合A={x²+3k²≥2k(2x-1},B={x|x²-(2x-1)k+k²≥0}且A∈B,试求k的取
设集合A={x²+3k²≥2k(2x-1},B={x|x²-(2x-1)k+k²≥0}且A∈B,试求k的取
答:
A={x²+3k²≥2k(2x-1}
x²+3k²>=2k(2x-1)
x²-4kx+3k²+2k>=0
(x-2k)²>=k²-2k
x>=2k+√(k²-2k)或者x=0
(x-k)²>=-k
x>=k+√(-k)或者x=0
-k>=0
2k-√(k²-2k)=k+√(-k)
所以:
k=k+√(-k)
√(k²-2k)>=-k+√(-k)
所以:
k=k²-2k√(-k)-k
所以:
k