Đáp án: $x\in\{\dfrac18,-\dfrac98\}$
Giải thích các bước giải:
Ta có:
$(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5})\cdot (2x+1)^2=\dfrac{10}{13}$
$\to \dfrac{4}{13}\cdot(\dfrac{6}{5}+\dfrac{2}{5})\cdot (2x+1)^2=\dfrac{10}{13}$
$\to \dfrac{4}{13}\cdot\dfrac{6+2}{5}\cdot (2x+1)^2=\dfrac{10}{13}$
$\to \dfrac{4}{13}\cdot\dfrac{8}{5}\cdot (2x+1)^2=\dfrac{10}{13}$
$\to \dfrac{32}{65}\cdot (2x+1)^2=\dfrac{10}{13}$
$\to (2x+1)^2=\dfrac{10}{13}:\dfrac{32}{65}$
$\to (2x+1)^2=\dfrac{25}{16}$
$\to 2x+1=\dfrac54\to 2x=\dfrac14\to x=\dfrac18$
Hoặc $2x+1=-\dfrac54\to 2x=-\dfrac94\to x=-\dfrac98$
