Question

若方程组｛x+y=4-m xy+m(x+y)=5 有实数解,求m的取值范围

Answer 1

｛x+y=4-m

xy+m(x+y)=5

xy+m(4-m)=5

x(4-m-x)+m(4-m)=5

-x^2+(4-m)x-(m^2-4m)-5=0

x^2+(m-4)x+(m^2-4m+5)=0

判别式=(m-4)^2-4*1*(m^2-4m+5)

=m^2-8m+16-4m^2+16m-20

=-3m^2+8m-4

方程组｛x+y=4-m xy+m(x+y)=5 有实数解

则有判别式>=0

-3m^2+8m-4>=0

3m^2-8m+4