Đáp án:
Giải thích các bước giải:
$\dfrac{1}{10}+\dfrac{1}{40}+...+\dfrac{1}{(3x+2)(3x+5)}=\dfrac{4}{25}$
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$⇒\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{(3x+2)(3x+5)}=\dfrac{12}{25}$
$ $
$⇒\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3x+2}-\dfrac{1}{3x+5}=\dfrac{12}{25}$
$ $
$⇒\dfrac{1}{2}-\dfrac{1}{3x+5}=\dfrac{12}{25}$
$ $
$⇒\dfrac{1}{3x+5}=\dfrac{1}{50}$
$ $
$⇒3x+5=50$
$⇒3x=45$
$⇒x=15$
