\(\frac{1}{2}a=2b\Leftrightarrow\frac{a}{2}=2b\Leftrightarrow a=4b\left(1\right)\)
\(\frac{9}{8}a=bk\Leftrightarrow\frac{9a}{8}=bk\Leftrightarrow9a=8bk\Leftrightarrow a=\frac{8bk}{9}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4b=\frac{8bk}{9}\)
\(\Leftrightarrow4b\cdot9=8bk\)
\(\Leftrightarrow36b=8bk\)
\(\Leftrightarrow\frac{36b}{8b}=k\)
\(\Leftrightarrow k=\frac{9}{2}\)
Vậy......
Có: \(\frac{1}{2}.a=2b\Rightarrow a=4b\)
\(\frac{9}{8}.a=kb\)
\(\Rightarrow\frac{9}{8}.4b=kb\)
\(\frac{9}{2}b=kb\)
\(\Leftrightarrow k=\frac{9}{2}\)\(\left(b\ne0\right)\)
Vậy \(k=\frac{9}{2}\)
