\(A=\dfrac{2mx-5}{x^2+n^2}\)
\(\Leftrightarrow Ax^2+An^2=2mx-5\)
\(\Leftrightarrow Ax^2-2mx+An^2+5=0\left(1\right)\)
A có cực trị khi (1) có nghiệm
\(\Leftrightarrow\Delta=4m^2-4A^2n^2-20A\ge0\)
\(\Leftrightarrow-A^2n^2-5A+m^2\ge0\left(1\right)\)
mà theo gt, ta có: \(\left\{{}\begin{matrix}A\ge-9\\A\le4\end{matrix}\right.\)
\(\Rightarrow\left(4-A\right)\left(A+9\right)\ge0\)
\(\Leftrightarrow-A^2-5A+36\ge0\left(2\right)\)
Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}n^2=1\\m^2=36\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=\pm1\\m=\pm6\end{matrix}\right.\)
Vậy \(m=\pm6;n=\pm1\)
\(-9\le A\le4\)
\(\dfrac{2mx-5}{x^2+n^2}\ge-9\Leftrightarrow2mx-5\ge-9\left(x^2+n^2\right)\)
<=>. 9x^2 +2mx +9n^2 -5 >=0
\(\Delta\le0\Leftrightarrow m^2-9\left(9n^2-5\right)\le0\)<=> m^2 -(9n)^2 +9.5 <=0 (a)
\(\dfrac{2mx-5}{x^2+n^2}\le4\Leftrightarrow2mx-5\le4\left(x^2+n^2\right)\)
<=>4x^2 -2mx +4n^2 +5 >=0
delta(x) <=0 <=>m^2 -4(4n^2 +5) <=0 <=> m^2 -(4n)^2 -4.5 <=0 (b)
đẳng thức xẩy ra m;n thỏa mãn hệ
m^2 -(9n)^2 +9.5 =0(1)
m^2 -(4n)^2 -4.5 =0 (2)
<=> [(9n) -(4n)][(9n) +(4n)]=4.5+9.5
<=> 5.13n^2 =13.5
<=>n^2 =1 => m^2 =9^2 -9.5 =9.4 =(2.3)^2
các cặp số m;n thảo mãn
(m;n) =(6;1);(-6;1);(6;-1);(6;1)
