来自李元科的问题
求值域y=x2-2x-3/x2-5x-6
求值域y=x2-2x-3/x2-5x-6
1回答
2020-12-27 04:21
求值域y=x2-2x-3/x2-5x-6
求值域y=x2-2x-3/x2-5x-6
推荐用Δ判别式法:y=(x²-2x-3)/(x²-5x-6)x²-2x-3=x²y-5xy-6y(1-y)x²+(5y-2)x+(6y-3)=0因为关于x的方程有解;所以(5y-2)²-4(1-y)(6y-3)≥049y²-56y+16≥0(7y-4)²≥0y∈R原...