Giải thích các bước giải:
Ta có :
$A=\dfrac{34}{7.13}+\dfrac{51}{13.22}+\dfrac{85}{22.37}+\dfrac{68}{37.49}$
$\to A=17(\dfrac{2}{7.13}+\dfrac{3}{13.22}+\dfrac{5}{22.37}+\dfrac{4}{37.49})$
$\to 3A=17(\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49})$
$\to 3A=17(\dfrac{13-7}{7.13}+\dfrac{22-13}{13.22}+\dfrac{37-22}{22.37}+\dfrac{49-37}{37.49})$
$\to 3A=17(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49})$
$\to 3A=17(\dfrac{1}{7}-\dfrac{1}{49})$
$\to A=\dfrac{17}{3}.(\dfrac{1}{7}-\dfrac{1}{49})$
Lại có :
$B=\dfrac{39}{7.16}+\dfrac{65}{16.31}+\dfrac{52}{31.43}+\dfrac{26}{43.49}$
$\to B=13(\dfrac{3}{7.16}+\dfrac{5}{16.31}+\dfrac{4}{31.43}+\dfrac{2}{43.49})$
$\to 3B=13(\dfrac{9}{7.16}+\dfrac{15}{16.31}+\dfrac{12}{31.43}+\dfrac{6}{43.49})$
$\to 3B=13(\dfrac{16-7}{7.16}+\dfrac{31-16}{16.31}+\dfrac{43-31}{31.43}+\dfrac{49-43}{43.49})$
$\to 3B=13(\dfrac{1}{7}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49})$
$\to 3B=13(\dfrac{1}{7}-\dfrac{1}{49})$
$\to B=\dfrac{13}{3}.(\dfrac{1}{7}-\dfrac{1}{49})$
$\to \dfrac{A}{B}=\dfrac{17}{13}$
