Giải thích các bước giải:

a.Khi $x=-3$

$\to B=\dfrac{-3+2}{3\cdot (-3)+2}=\dfrac17$

b.Ta có:

$A=\dfrac1{x+2}-\dfrac{2x}{4-x^2}+\dfrac3{x-2}$

$\to A=\dfrac1{x+2}+\dfrac{2x}{x^2-4}+\dfrac3{x-2}$

$\to A=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}$

$\to A=\dfrac{3\left(x+2\right)+3x-2}{\left(x+2\right)\left(x-2\right)}$

$\to A=\dfrac{6x+4}{\left(x+2\right)\left(x-2\right)}$

$\to M=\dfrac{6x+4}{\left(x+2\right)\left(x-2\right)}\cdot \dfrac{x+2}{3x+2}$

$\to M=\dfrac2{x-2}$

c.Ta có:

$N=\dfrac2{x-2}\cdot (x^3-x^2-2x)=\dfrac2{x-2}\cdot x(x+1)(x-2)=2x(x+1)=2(x^2+x)=2((x+\dfrac12)^2-\dfrac14)=2(x+\dfrac12)^2-\dfrac12\ge-\dfrac12$

Dấu = xảy ra khi $x+\dfrac12=0\to x=-\dfrac12$