Question

[Nguyễn Trần Gia Huy]

Tìm các số tự nhiên a; b thoả mãn điều kiện : 11/ 17 < a b < 23/ 29 và 8b-9a=31

Answer 1

[꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...]

Giải:

Ta biết: \(\dfrac{11}{17}< \dfrac{a}{b}< \dfrac{23}{29}\) và \(8b-9a=31\) \(\left(a;b\in N\right)\)

Theo đề bài: \(8b-9a=31\) 

\(\Rightarrow b=\dfrac{31+9a}{8}=\dfrac{32-1+8a+a}{8}=\left[\left(4+a\right)+\dfrac{a-1}{8}\right]\in N\) 

\(\Leftrightarrow\dfrac{a-1}{8}\in N\) 

\(\Leftrightarrow\left(a-1\right)⋮8\) 

\(\Leftrightarrow a=8k+1\left(k\in N\right)\) 

Khi đó:

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}11.\left(9k+5\right)< 17.\left(8k+1\right)\Leftrightarrow k>1\\29.\left(8k+1\right)< 23.\left(9k+5\right)\Leftrightarrow k< 4\end{matrix}\right.\) 

\(\Rightarrow1< k< 4\)

\(\Rightarrow k\in\left\{2;3\right\}\) 

Với \(\left[{}\begin{matrix}k=2\Rightarrow\left\{{}\begin{matrix}a=17\\b=23\end{matrix}\right.\\k=3\Rightarrow\left\{{}\begin{matrix}a=25\\b=32\end{matrix}\right.\end{matrix}\right.\) 

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

Answer 2

[Phạm An Tường]

Giải:

Ta biết: $\genfrac{}{}{0ex}{}{11}{17}<\genfrac{}{}{0ex}{}{a}{b}<\genfrac{}{}{0ex}{}{23}{29}$ và $8b-9a=31$ $\left(a;b\in N\right)$

Theo đề bài: $8b-9a=31$ 

$\Rightarrow b=\genfrac{}{}{0ex}{}{31+9a}{8}=\genfrac{}{}{0ex}{}{32-1+8a+a}{8}=\left[\left(4+a\right)+\genfrac{}{}{0ex}{}{a-1}{8}\right]\in N$ 

$\iff \genfrac{}{}{0ex}{}{a-1}{8}\in N$ 

$\iff \left(a-1\right)⋮8$ 

$\iff a=8k+1\left(k\in N\right)$ 

Khi đó:

$b=\genfrac{}{}{0ex}{}{31+9.\left(8k+1\right)}{8}=9k+5$ 

$\Rightarrow \genfrac{}{}{0ex}{}{11}{17}<\genfrac{}{}{0ex}{}{8k+1}{9k+5}<\genfrac{}{}{0ex}{}{23}{29}$ 

$\iff \{\begin{array}{c}11.\left(9k+5\right)<17.\left(8k+1\right)\iff k>1\\ 29.\left(8k+1\right)<23.\left(9k+5\right)\iff k<4\end{array}$ 

$\Rightarrow 1<k<4$

$\Rightarrow k\in \left\{2;3\right\}$ 

Với $[\begin{array}{c}k=2\Rightarrow \{\begin{array}{c}a=17\\ b=23\end{array}\\ k=3\Rightarrow \{\begin{array}{c}a=25\\ b=32\end{array}\end{array}$ 

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