Giải thích các bước giải:
a.Ta có:
$A=\dfrac{1}{x^2+x}+\dfrac1{x+1}$
$\to A=\dfrac{1}{x(x+1)}+\dfrac{x}{x(x+1)}$
$\to A=\dfrac{1+x}{x(x+1)}$
$\to A=\dfrac1x$
b.Ta có:
$P=A:B$
$\to P=\dfrac1x:\dfrac2{x+1}$
$\to P=\dfrac{x+1}{2x}$
c.Để $P=3$
$\to \dfrac{x+1}{2x}=3$
$\to x+1=6x$
$\to x=\dfrac15$
d.Ta có $C=2x^2\cdot P$
$\to C=2x^2\cdot \dfrac{x+1}{2x}$
$\to C=x(x+1)$
$\to C=x^2+x$
$\to C=(x+\dfrac12)^2-\dfrac14\ge -\dfrac14$
Dấu = xảy ra khi $x+\dfrac12=0\to x=-\dfrac12$
e.Để $P>\dfrac12$
$\to \dfrac{x+1}{2x}>\dfrac12$
$\to \dfrac{x+1}{x}>1$
$\to 1+\dfrac1x>1$
$\to \dfrac1x>0$
$\to x>0$