y=(mx^2+4√3x+n)/(x^2+1) (m-y)x^2+4√3x+n-y=0 上方程未知数为x的判别式△≥0 即(4√3)^2-4(m-y)*(n-y)≥0 y^2-(m+n)y+mn-12≤0 [m+n-√(m^2+n^2-2mn+48)]/2≤y≤[m+n+√(m^2+n^2-2mn+48)]/2 已知y=(mx^2+4√3x+n)/(x^2+1)(m<n)的值域为[-1,7],可知 [m+n-√(m^2+n^2-2mn+48)]/2=-1.(1) [m+n+√(m^2+n^2-2mn+48)]/2=7.(2) (1)+(2) m+n=6.(3) (2)-(1) √(m^2+n^2-2mn+48)=8 m^2+n^2-2mn+48=64 (n-m)^2=16 m<n -m+n=4.(4) (3)+(4) n=5 (3)-(4) m=1 答:m=1,n=5