y=(3x-2)^(1/2)+(2-3x)^(-1/3)的定义域_________________________________________________
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y=(3x-2)^(1/2)+(2-3x)^(-1/3)的定义域_________________________________________________y=(3x-2)^(1/2)+(2-3
y=(3x-2)^(1/2)+(2-3x)^(-1/3)的定义域_________________________________________________
y=(3x-2)^(1/2)+(2-3x)^(-1/3)的定义域_________________________________________________
y=(3x-2)^(1/2)+(2-3x)^(-1/3)的定义域_________________________________________________
(3x-2)^(1/2)=√(3x-2)
所以3x-2>=0
x>=2/3
(2-3x)^(-1/3)=1/(2-3x)^(1/3)
所以分母2-3x≠0
x不等于2/3
所以定义域是(2/3,+∞)
3x-2≥0
2-3x≠0
解得:x>2/3
所以:定义域为x∈(2/3,+∞)
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