Đáp án: $x=15$
Giải thích các bước giải:
Ta có :
$\dfrac7x+\dfrac4{5\cdot9}+\dfrac4{9\cdot13}+...+\dfrac4{41\cdot45}=\dfrac{29}{45}$
$\to\dfrac7x+\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9\cdot13}+...+\dfrac{45-41}{41\cdot45}=\dfrac{29}{45}$
$\to\dfrac7x+\dfrac15-\dfrac19+\dfrac19-\dfrac1{13}+...+\dfrac1{41}-\dfrac1{45}=\dfrac{29}{45}$
$\to\dfrac7x+\dfrac15-\dfrac1{45}=\dfrac{29}{45}$
$\to\dfrac7x+\dfrac8{45}=\dfrac{29}{45}$
$\to\dfrac7x=\dfrac{21}{45}$
$\to\dfrac7x=\dfrac7{15}$
$\to x=15$
