Giải thích các bước giải:

a.Ta có:

Đặt $(x+\dfrac1x)^2=t, t\ge 0$

$\to x^2+2x\cdot\dfrac1x+\dfrac1{x^2}=t$

$\to x^2+\dfrac1{x^2}+2=t$ (suy ra $t>2$)

$\to x^2+\dfrac1{x^2}=t-2$

$\to A=\sqrt{(t-2)^2+2t-3}:(x^2-x+1)$

$\to A=\sqrt{t^2-2t+1}:(x^2-x+1)$

$\to A=\sqrt{(t-1)^2}:(x^2-x+1)$

$\to A=(t-1):(x^2-x+1)$ vì $t>2$

$\to A=((x+\dfrac1x)^2-1):(x^2-x+1)$

$\to A=(x^2+\dfrac1{x^2}+1):(x^2-x+1)$

$\to A=\dfrac{x^4+x^2+1}{x^2}:(x^2-x+1)$

$\to A=\dfrac{x^4+2x^2+1-x^2}{x^2}:(x^2-x+1)$

$\to A=\dfrac{(x^2+1)^2-x^2}{x^2}:(x^2-x+1)$

$\to A=\dfrac{(x^2+1-x)(x^2+1+x)}{x^2}:(x^2-x+1)$

$\to A=\dfrac{x^2+x+1}{x^2}$

b.Ta có:

$A=\dfrac{x^2+x+1}{x^2}$

$\to A=1+\dfrac1x+\dfrac1{x^2}$

$\to A=\dfrac34+(\dfrac1x+\dfrac12)^2\ge\dfrac34$

$\to GTNN=\dfrac34$

$\to \dfrac1x+\dfrac12=0$

$\to x=-2$