Question

Tìm \(\dfrac{a}{b}\) thỏa mãn:

\(\dfrac{4}{7}< \dfrac{a}{b}< \dfrac{2}{3}\) và 7a + 4b = 1994

Answer 1

Có: \(\frac{a}{b}\left(a,b\in N,b\ne0\right)\)

\(\frac{a}{b}>\frac{4}{7}\)

\(\Rightarrow7a>4b\)

\(\Leftrightarrow8b< 1994\)

\(\Leftrightarrow b< 249\)

\(7a>4b\)

\(\Leftrightarrow14a>1994\)

\(\Leftrightarrow a>142\)

Có: \(\frac{a}{b}< \frac{2}{3}\)

\(\Rightarrow3a< 2b\)

\(\Leftrightarrow6a+7a< 4b+7a\)

\(\Leftrightarrow13a< 1994\)

\(\Leftrightarrow a< 154\)

Có:\(3a< 2b\)

\(\Leftrightarrow6a+a+4b< 8b+a\)

\(\Leftrightarrow1994< 8b+a\)

mà a=\(\frac{1994-4b}{7}\)

\(8b+a=8b+\frac{1994-4b}{7}>1994\)

\(\Leftrightarrow56b+1994-4b>13958\)

\(\Leftrightarrow b>230\)

Vậy \(\frac{4}{7}< \frac{a}{b}< \frac{2}{3}\Leftrightarrow a,b\in N;142< a< 154;230< b< 249\)

Nguyễn Việt Lâm Bài này có cần tìm cụ thể ko?