ĐKXĐ: $x\ne ±2$
a/ Với $x=\dfrac{1}{2}$ (thỏa mãn điều kiện)
$→A=\dfrac{\dfrac{1}{2}-2}{\left(\dfrac{1}{2}\right)^2+1}\\=\dfrac{-\dfrac 3 2}{\dfrac 5 4}\\=-\dfrac{6}{5}$
b/ $B=\dfrac{3}{x-2}+\dfrac{6-5x}{4-x^2}+\dfrac{2x}{x+2}\\=\dfrac{3(x+2)}{(x-2)(x+2)}-\dfrac{6-5x}{(x-2)(x+2)}+\dfrac{2x(x-2)}{(x-2)(x+2)}\\=\dfrac{3x+6-6+5x+2x^2-4x}{(x-2)(x+2)}\\=\dfrac{2x^2+4x}{(x-2)(x+2)}\\=\dfrac{2x(x+2)}{(x-2)(x+2)}\\=\dfrac{2x}{x-2}$
$→$ ĐPCM
c/ $P=A.B\\=\dfrac{x-2}{x^2+1}.\dfrac{2x}{x-2}\\=\dfrac{2x}{x^2+1}$
$P\le 1\\↔\dfrac{2x}{x^2+1}\le 1\\↔\dfrac{2x}{x^2+1}-1\le 0\\↔\dfrac{2x-x^2-1}{x^2+1}\le 0\\↔-\dfrac{x^2-2x+1}{x^2+1}\le 0\\↔\dfrac{(x-1)^2}{x^2+1}\ge 0$
Nhận thấy: $\begin{cases}(x-1)^2\ge 0\\x^2+1>0\end{cases}$
$→P≥0∀x\\→P≤1∀x$
Vậy $P≤1∀x$
