Đáp án: $A=1$
Giải thích các bước giải:
Ta có :
$A=\dfrac{\dfrac13-\dfrac17-\dfrac1{13}}{\dfrac23-\dfrac2{13}-\dfrac27}\cdot\dfrac{\dfrac13-0.25+0.2}{1\dfrac16-0.875+0.7}+\dfrac67$
$\to A=\dfrac{\dfrac13-\dfrac17-\dfrac1{13}}{\dfrac23-\dfrac27-\dfrac2{13}}\cdot\dfrac{\dfrac13-\dfrac14+\dfrac15}{\dfrac76-\dfrac78+\dfrac7{10}}+\dfrac67$
$\to A=\dfrac{\dfrac13-\dfrac17-\dfrac1{13}}{2(\dfrac13-\dfrac17-\dfrac1{13})}\cdot\dfrac{\dfrac13-\dfrac14+\dfrac15}{\dfrac72(\dfrac13-\dfrac14+\dfrac1{5})}+\dfrac67$
$\to A=\dfrac{1}{2}\cdot\dfrac{1}{\dfrac72}+\dfrac67$
$\to A=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac67$
$\to A=\dfrac{1}{7}+\dfrac67$
$\to A=1$
