Tìm giá trị nhỏ nhất và giá trị lớn nhất của biểu thức : $Q=\frac{(x+y)(1-xy)}{(1+x^2)(1+y)^2}$
Với $\forall k \in R,$ta có $Q+k=\frac{(x+y)(1-xy)+k(1+x^2)(1+y^2)}{(1+x^2)(1+y)^2}$
 $=\frac{[k+(1+y^2)-y]x^2+(1-y^2)x+k(1+y^2)+y}{(1+x^2)(1+y^2)}$
Gọi $\Delta =(1-y^2)^2-4[k(1+y^2)-y][k(1+y^2)+y]=(1-4k^2)(1+y^2)$
Suy ra $\Delta =0\Leftrightarrow k= \pm \frac{1}{2}$
$\bullet$ với $k=\frac{1}{2}$ có $Q+\frac{1}{2}=\frac{(1+y^2-2y)x^2+2(1-y^2)x+1+y^2+2y}{2(1+x^2)(1+y^2)}$
$=\frac{(1-y)^2+x^2+2((1-y^2)x+(1+y)^2}{2(1+x^2)(1+y^2)} =\frac{(1+x+y-xy)^2}{2(1+x^2)(1+y^2)} \geq 0$
$ \Rightarrow Q\geq \frac{-1}{2}$ Dấu đẳng thức có khi và chỉ khi $x+y=xy-1$
$\Rightarrow$ GTNN của $Q$ là $Q = \frac{-1}{2}$
$\bullet$ Với $k=\frac{-1}{2}$ có $Q-\frac{1}{2}=\frac{-(1+y^2+2y)x^2+2(1-y)^2x-(1+y^2-2y)}{2(1+x^2)1+y^2)}$
$=\frac{-(1+y)^2x^2+2(1-y^2)x-(1-y)^2}{2(1+x^2)(1+y^2)}=-\frac{(1-x-y-xy)^2}{2(1+x^2)(1+y^2)} \leq 0$
$\Rightarrow Q \leq \frac{1}{2}$.Dấu đẳng thức có khi và chỉ khi $x+y=1-xy$
$\Rightarrow $ GTNN của $Q$ là $Q=\frac{1}{2}$      

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