Giải thích các bước giải:
a.Ta có : $A=\dfrac{3-4x}{x^2+1}$
$\to A+1=\dfrac{3-4x}{x^2+1}+1$
$\to A+1=\dfrac{3-4x+x^2+1}{x^2+1}$
$\to A+1=\dfrac{x^2-4x+4}{x^2+1}$
$\to A+1=\dfrac{(x-2)^2}{x^2+1}$
Vì $(x-2)^2\ge 0, x^2+1>0\quad\forall x$
$\to A+1=\dfrac{(x-2)^2}{x^2+1}\ge 0$
$\to A\ge -1$
$\to GTNN_A=-1\to x=2$
b.Ta có :
$A=\dfrac{3-4x}{x^2+1}$
$\to 4-A=4-\dfrac{3-4x}{x^2+1}$
$\to 4-A=\dfrac{4(x^2+1)-(3-4x)}{x^2+1}$
$\to 4-A=\dfrac{4x^2+4x+1}{x^2+1}$
$\to 4-A=\dfrac{(2x+1)^2}{x^2+1}$
Vì $(2x+1)^2\ge 0 ,x^2+1>0\quad\forall x$
$\to 4-A=\dfrac{(2x+1)^2}{x^2+1}\ge 0$
$\to A\le 4$
$\to GTLN_A=4\to x=-\dfrac12$
