Giải thích các bước giải:

d/ $\dfrac{3x+2}{-32}=\dfrac{-2}{2+3x}$

$\text{ĐKXĐ: $x \neq -\dfrac{2}{3}$}$

$⇔ (3x+2)^2=(-32).(-2)$

$⇔ (3x+2)^2=64$

$⇔\left[ \begin{array}{l}3x+2=8\\3x+2=-8\end{array} \right.$

$⇔ \left[ \begin{array}{l}3x=6\\3x=-10\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=2(TM)\\x=-\dfrac{10}{3}(TM)\end{array} \right.$

g/ $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.....+\dfrac{2}{x(x+1)}=\dfrac{2}{2013}$

$\text{ĐKXĐ: $x \neq 0$ và $x \neq -1$}$

$⇔ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+....+\dfrac{2}{x(x+1)}=\dfrac{2}{2013}$

$⇔ \dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+....+\dfrac{2}{x(x+1)}=\dfrac{2}{2013}$

$⇔ 2(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1})=\dfrac{2}{2013}$

$⇔ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2}{2013}.\dfrac{1}{2}$

$⇔ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1}{2013}$

$⇔ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1}{2013}=\dfrac{2011}{4026}$

$⇔ x+1=\dfrac{4026}{2011}$

$⇔ x=\dfrac{2015}{2011}(TM)$

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Đáp án:

$d) \dfrac{3x+2}{-32} = \dfrac{-2}{2+3x}$

$\text{ĐKXĐ : x $\neq$ -$\dfrac{2}{3}$ }$

$⇔\dfrac{(3x+2)(2+3x)}{-32(2+3x)} = \dfrac{-2.(-32)}{-32(2+3x)}$

$⇒ (3x+2)(2+3x) = 64$

$⇔6x+9x^2+4+6x = 64$

$⇔ 9x^2+6x+6x+4-64 =0$

$⇔9x^2+12x-60=0$

$⇔ 9x^2-18x+30x-60=0$

$⇔9x(x-2)+30(x-2)=0$

$⇔(x-2)(9x+30)=0$

⇔\(\left[ \begin{array}{l}x-2=0\\9x+30=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=2\\x=-\dfrac{10}{3}\end{array} \right.\) 

$\text{Vậy x ∈ {-$\dfrac{10}{3}$ ; 2}}$

$g) \dfrac{1}{3} +\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x(x+1)} = \dfrac{2}{2013}$

$⇔\dfrac{2}{6} +\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x(x+1)} =\dfrac{2}{2013}$

$⇔2(\dfrac{1}{2} -\dfrac{1}{3} +\dfrac{1}{3} -\dfrac{1}{4} +\dfrac{1}{4} -\dfrac{1}{5} +...+\dfrac{1}{x} -\dfrac{1}{x+1} = \dfrac{2}{2013} $

$⇔\dfrac{1}{2} -\dfrac{1}{x+1}=\dfrac{2}{2013} .\dfrac{1}{2}$

$⇔\dfrac{1}{2} -\dfrac{1}{x+1} =\dfrac{1}{2013}$

$⇔\dfrac{1}{x+1} =\dfrac{1}{2} -\dfrac{1}{2013}$

$⇔\dfrac{1}{x+1}= \dfrac{2011}{4026}$

$⇔x+1 = \dfrac{4026}{2011}$

$⇔x=\dfrac{4026}{2011} -1$

$⇔x=\dfrac{2015}{2011}$

$\text{Vậy x =$\dfrac{2015}{2011}$}$