求Y²=(X²+4X+3)/(X²+x-6)的值域用判别式法,求详解
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求Y²=(X²+4X+3)/(X²+x-6)的值域用判别式法,求详解
求Y²=(X²+4X+3)/(X²+x-6)的值域
用判别式法,求详解
求Y²=(X²+4X+3)/(X²+x-6)的值域用判别式法,求详解
令a=y²
ax²+ax-6a=x²+4x+3
(a-1)x²+(a-4)x-(6a+3)=0
△=a²-8a+16+24a²-12a-12>=0
25a²-20a+4>=0
(5a-2)²>=0
成立
若a=1
则x²+4x+3=x²+x-6
x=-3,分母为0,不成立
所以y²≠1
所以y≠±1