mn giúp mình với ạ. T^T

\(\frac{5a+7b}{6a+5b}=\frac{29}{28}\Rightarrow28\left(5a+7b\right)=29\left(6a+5b\right)\Rightarrow140a+196b=174a+145b\)

\(\Rightarrow\left(140a+196b\right)-\left(174a+145b\right)=0\Rightarrow140a+196b-174a-145b=0\)

\(\Rightarrow140a-174a+196b-145b=0\Rightarrow\left(-34\right)a+51b=0\)

\(\Rightarrow51b=34a\Rightarrow\frac{a}{51}=\frac{b}{34}\)

G/S:\(\frac{a}{51}=\frac{b}{34}=k\Rightarrow a=51k;b=34k\)

                                        \(\frac{5a+7b}{6a+5b}=\frac{29}{28}\Rightarrow\frac{5.51k+7.34k}{6.51k+5.34k}=\frac{255k+238k}{306k+170k}=\frac{493k}{476k}=\frac{29k}{28k}=\frac{29}{28}\Rightarrow k=1\)

\(\Rightarrow a=51.1=51;b=34.1=34\)

D/S:  ........

Phần cuối bài làm chỉ cần ra a/b = 51/34. Mà (a,b) =1 suy ra a = 51 và b = 34. 

\(28\left(5a+7b\right)=29\left(6a+5b\right)\Leftrightarrow140a+196b=174a+145b\Leftrightarrow51b=34a\Leftrightarrow\frac{a}{b}=\frac{51}{34}=\frac{3}{2}\)

Vì (a;b) =1

=> a =3 ; b =2

Để thỏa mãn điều kiện trên thì :

 ( 5a + 7b ) x 28 = 29 x ( 6a + 5b )

140a + 196b = 174a + 145b

=> 34a = 41b

=> a = 41 ; b = 34

(5a+7b).28=29.(6a+5b) =>140a+196b=174a+145b =>2a=3b =>a/b=3/2 => a=3k; b=2k mà (a,b)=1 nên a=3; b=2

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

\(\Leftrightarrow29\left(5a+7b\right)=28\left(6a+5b\right)\)

\(\Leftrightarrow145a+203b=168a+140b\)

\(\Leftrightarrow63b=23a\)

\(\Leftrightarrow\frac{a}{b}=\frac{63}{23}\)

Mà \(\left(a;b\right)=1\) nên \(a=63;b=23\)