=>(5a+7b)*28=(6a+5b)*29 
140a+196b=174a+145b 
140a-174a=-196b+145b 
-34a=-51b 
=>-34/-51=b/a=2/3=b/a=> b=2;a=3\

tk mk

Nhân chéo 2 vế ta có: 140a+196b = 174a+145b

                                 \(\Rightarrow\) 51b = 34a

                                 \(\Rightarrow\) \(\frac{a}{b}=\frac{51}{34}=\frac{3}{2}\) 

Vậy a = 3 ; b = 2

Giải:

Theo đề bài ta có:

\(8b-9a=31\Rightarrow b=\dfrac{31+9a}{8}\)

\(=\dfrac{32-1+8a+a}{8}=\left[\left(4+a\right)+\dfrac{a-1}{8}\right]\) \(\in N\)

\(\Rightarrow\dfrac{a-1}{8}\in N\Leftrightarrow\left(a-1\right)⋮8\Rightarrow a=8k+1\left(k\in N\right)\)

Khi đó: \(b=\dfrac{31+9\left(8k+1\right)}{8}=9k+5\)

\(\Rightarrow\dfrac{11}{17}< \dfrac{8k+1}{9k+5}< \dfrac{23}{29}\)

\(\Rightarrow11\left(9k+5\right)< 17\left(8k+1\right)\Rightarrow37k>38\) \(\Rightarrow k>1\left(1\right)\)

Và \(29\left(8k+1\right)< 23\left(9k+5\right)\Rightarrow25k< 86\) \(\Rightarrow k< 4\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow1< k< 4\Leftrightarrow k\in\left\{2;3\right\}\)

Ta xét 2 trường hợp:

Trường hợp 1: Nếu \(k=2\)

\(\Rightarrow\left\{{}\begin{matrix}a=8k+1\\b=9k+5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=8.2+1\\b=9.2+5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=17\\b=23\end{matrix}\right.\)

Trường hợp 2: Nếu \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=8k+1\\b=9k+5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=8.3+1\\b=9.3+5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=25\\b=32\end{matrix}\right.\)

Vậy \(\left(a,b\right)=\left(17;23\right);\left(25;32\right)\)

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

=> 5a + 7b = 28

va 6a + 5b = 29

=> a = 3 va b = 2

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

=> 5b + 7b = 28

và 6a + 5b = 29

=> a = 3 và b = 2

=>5a+7b=28

va 6a+5b=29

a=3 ,b=2