Giải thích các bước giải:

a.ĐKXĐ: $x\ne 0$

Ta có:

$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac3{x(x^4+x^2+1)}$

$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac3{x(x^4+2x^2+1-x^2)}$

$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac3{x((x^2+1)^2-x^2)}$

$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac3{x(x^2+1-x)(x^2+1+x)}$

$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac3{x(x^2-x+1)(x^2+x+1)}$

$\to (x+1)(x^2-x+1)-(x-1)(x^2+x+1)=\dfrac3x$

$\to (x^3+1)-(x^3-1)=\dfrac3x$

$\to 2=\dfrac3x$

$\to x=\dfrac32$

b.ĐKXĐ: $x\not\in\{-4, -5, -6, -7\}$

Ta có:

$\dfrac1{x^2+9x+20}+\dfrac1{x^2+11x+30}+\dfrac1{x^2+13x+42}=18$

$\to \dfrac1{(x+4)(x+5)}+\dfrac1{(x+5)(x+6)}+\dfrac1{(x+6)(x+7)}=18$

$\to \dfrac1{x+4}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+6}+\dfrac1{x+6}-\dfrac1{x+7}=18$

$\to \dfrac1{x+4}-\dfrac1{x+7}=18$

$\to x+7-\left(x+4\right)=18\left(x+4\right)\left(x+7\right)$

$\to 3=18x^2+198x+504$

$\to 8x^2+198x+501=0$

$\to x=\dfrac{-33\pm\sqrt{87}}{6}$

c.Đặt $x^2-2x+2=t\to t=(x-1)^2+1\ge 1$

Ta có:

$\dfrac1t+\dfrac2{t+1}=\dfrac6{t+2}$

$\to \left(t+1\right)\left(t+2\right)+2t\left(t+2\right)=6t\left(t+1\right)$

$\to 3t^2+7t+2=6t^2+6t$

$\to -3t^2+t+2=0$

$\to -(3t+2)(t-1)=0$

$\to t=1$ vì $t\ge 1$

$\to x^2-2x+2=1$

$\to x^2-2x+1=0$

$\to (x-1)^2=0$

$\to x=1$