Question

Tìm các số tự nhiên a và b để thỏa mãn \(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)vàƯCLN(a,b)=1

Answer 1

\(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)

\(\Rightarrow28\left(5a+7b\right)=29\left(6a+5b\right)\)

\(\Rightarrow140a+196b=174a+145b\)

\(\Rightarrow140a=174a+145b-196b\)

\(\Rightarrow140a=174a-51b\)

\(\Rightarrow34a=51b\)

\(\Rightarrow2a=3b\)

Mà \(\left(a,b\right)=1\) nên \(a=3;b=2\)

Answer 2

\(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)

\(\Rightarrow140a+196b=174a+145b\)

\(\Rightarrow51b=34a\)

\(\Rightarrow3b=2a\)

Do \(\left(a;b\right)=1\)

\(\Rightarrow a=3,b=2\)

Vậy...