7/x + 4/5 x 9 =4 / 9x 13 +….+ 4/ 41.45 = 29/45

Đáp án: $x=15$

Giải thích các bước giải:

Ta có :
$\dfrac7x+\dfrac4{5\cdot9}+\dfrac4{9\cdot13}+…+\dfrac4{41\cdot45}=\dfrac{29}{45}$

$\to\dfrac7x+\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9\cdot13}+…+\dfrac{45-41}{41\cdot45}=\dfrac{29}{45}$

$\to\dfrac7x+\dfrac15-\dfrac19+\dfrac19-\dfrac1{13}+…+\dfrac1{41}-\dfrac1{45}=\dfrac{29}{45}$

$\to\dfrac7x+\dfrac15-\dfrac1{45}=\dfrac{29}{45}$

$\to\dfrac7x+\dfrac8{45}=\dfrac{29}{45}$

$\to\dfrac7x=\dfrac{21}{45}$

$\to\dfrac7x=\dfrac7{15}$

$\to x=15$