Question

[Khang Trần]

Tìm \(a,b\in N\) để\(\dfrac{11}{17}\)<\(\dfrac{a}{b}\)<\(\dfrac{23}{29}\)và 8b-9a=31

Answer 1

[Citii?]

Ta có: \(8b-9a=31\)

\(\Rightarrow8b=31+9a\)

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

\(\Rightarrow b=\dfrac{32+8a+a-1}{8}\)

\(\Rightarrow b=\dfrac{8\cdot\left(4+a\right)+a-1}{8}\)

\(\Rightarrow b=4+a+\dfrac{a-1}{8}\)

Để \(b\in N\) thì:

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

\(\Rightarrow a-1⋮8\)

\(\Rightarrow a-1=8k\left(k\in N\right)\)

\(\Rightarrow a=8k+1\)

Khi đó: \(b=4+8k+1+\dfrac{8k+1-1}{8}\)

\(\Rightarrow b=5+8k+\dfrac{8k}{8}\)

\(\Rightarrow b=5+8k+k\)

\(\Rightarrow b=5+9k\)

Mặt khác: \(\dfrac{11}{17}< \dfrac{a}{b}< \dfrac{23}{29}\)

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

Xét: \(\dfrac{11}{17}< \dfrac{8k+1}{5+9k}\)

\(\Rightarrow11\left(5+9k\right)< 17\left(8k+1\right)\)

\(\Rightarrow55+99k< 136k+17\)

\(\Rightarrow136k-99k>55-17\)

\(\Rightarrow37k>38\)

\(\Rightarrow k>\dfrac{38}{37}\left(1\right)\)

Xét: \(\dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)

\(\Rightarrow29\left(8k+1\right)< 23\left(5+9k\right)\)

\(\Rightarrow232k+29< 115+207k\)

\(\Rightarrow232k-207k< 115-29\)

\(\Rightarrow25k< 86\)

\(\Rightarrow k< \dfrac{86}{25}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{38}{27}< k< \dfrac{86}{25}\)

Mà \(k\in N\)

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

\(+,\) \(k=2\).

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot2+1=17\\b=5+9\cdot2=23\end{matrix}\right.\)

\(+,\) \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot3+1=25\\b=5+9\cdot3=32\end{matrix}\right.\)

\(\Rightarrow\) Vậy \(\left(a;b\right)\in\left\{\left(17;23\right),\left(25;32\right)\right\}\)