Tìm GTLN và GTNN của biểu thức $\frac{x^{2}+6x+1}{x^{2}+1}$
$\frac{x^{2}+6x+1}{x^{2}+1}=1+\frac{6x}{x^{2}+1}$
ta tìm GTLN,NN của $B=\frac{6x}{x^{2}+1}$
xét $B+3=3\frac{(x+1)^{2}}{x^{2}+1}\geqslant 0\Rightarrow GTNN$
$B-3\leqslant 0\Rightarrow GTLN$

Đặt $\frac{x^2+6x+1}{x^2+1}=y\Rightarrow (y-1).x^2-6x+y-1=0$
Ta xét $y=1$ thì $x = 0$
Xét $y\neq 1$ : pt trên có nghiệm $\Leftrightarrow \Delta' \geq 0\Leftrightarrow 3^2-(y-1)^2 \geq 0\Leftrightarrow (4-y)(y+2)\geq 0\Leftrightarrow -2\leq y\leq 4\Rightarrow $ Biểu thức có giá trị Max là 4, Min là -2
Đặt $A=\frac{x^{2}+ 6x +1}{x^{2}+1}$
$\Rightarrow Ax^{2} + A = x^{2} + 6x +1$
$\Rightarrow (A-1)x^{2} - 6x +(A-1) = 0$
$\Delta = 36 - 4(A - 1)^{2} = -4A^{2} + 8A + 32$
$ \Delta \geq 0 \Rightarrow -4A^{2} + 8A + 32 \geq  0$
Giải bất phương trình ta được $-2 \leq A \leq 4$
Vậy: GTLN là $4$ khi $ x = 1$
        GTNN là $ -2$ khi $x= -1$

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