Giải thích các bước giải:

a.Khi $x=3\to A=\dfrac{3^2}{3+3}=\dfrac32$

b.Ta có:

$B=\dfrac{3x}{x-1}-\dfrac{2x}{x-1}+\dfrac{x-3}{1-x^2}$

$\to B=\dfrac{x}{x-1}-\dfrac{x-3}{x^2-1}$ 

$\to B=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-3}{\left(x-1\right)\left(x+1\right)}$

$\to B=\dfrac{x\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}$

$\to B=\dfrac{x^2+3}{\left(x-1\right)\left(x+1\right)}$

$\to$Xem lại đề

c.Ta có:

$P=\dfrac{x^2}{x+3}+\dfrac{x+3}{x-1}=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=(x+1)+\dfrac1{x-1}=(x-1)+\dfrac1{x-1}+2\ge 2\sqrt{(x-1)\cdot \dfrac1{x-1}}+2=4$

Dấu = xảy ra khi $x-1=\dfrac1{x-1}\to x=2$ vì $x>1$