设集合m=﹛x|x=x²-3x+2=0﹜,n=﹛x|2x²-ax+2=0﹜,若m并n等于m,求a得取值集合
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![设集合m=﹛x|x=x²-3x+2=0﹜,n=﹛x|2x²-ax+2=0﹜,若m并n等于m,求a得取值集合](/uploads/image/z/2618067-3-7.jpg?t=%E8%AE%BE%E9%9B%86%E5%90%88m%3D%EF%B9%9Bx%7Cx%3Dx%26%23178%3B-3x%2B2%3D0%EF%B9%9C%2Cn%3D%EF%B9%9Bx%7C2x%26%23178%3B-ax%2B2%3D0%EF%B9%9C%2C%E8%8B%A5m%E5%B9%B6n%E7%AD%89%E4%BA%8Em%2C%E6%B1%82a%E5%BE%97%E5%8F%96%E5%80%BC%E9%9B%86%E5%90%88)
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设集合m=﹛x|x=x²-3x+2=0﹜,n=﹛x|2x²-ax+2=0﹜,若m并n等于m,求a得取值集合
设集合m=﹛x|x=x²-3x+2=0﹜,n=﹛x|2x²-ax+2=0﹜,若m并n等于m,求a得取值集合
设集合m=﹛x|x=x²-3x+2=0﹜,n=﹛x|2x²-ax+2=0﹜,若m并n等于m,求a得取值集合
m并n等于m
即
n包含于m
m:x²-3x+2=0,(x-1)(x-2)=0
x=1或x=2
m={1,2}
n={1}
2-a+2=0
a=4
2x²-4x+2=0,x1,2=1,可以;
n={2}
x=2代入,得
8-2a+2=0
a=5
错;
n=∅
a²-4×2×2<0
a²<16
-4<a<4
所以
-4<a≤4.