Question

مطلوب است محاسبه : $$ \frac{9a+b}{9c+d} =?$$ در صورتی که: $$ \frac{6a+b}{6c+d} = \frac{5a+b}{5c+d} \wedge a,b,c,d > 0 \wedge \frac{7a+b}{7c+d} =8$$

Answer 1

$$ \frac{6a+b}{6c+d} = \frac{5a+b}{5c+d} \Longrightarrow \frac{6a+b}{5a+b} = \frac{6c+d}{5c+d} \Longrightarrow \frac{a}{5a+b} +1= \frac{c}{5c+d} +1 \Longrightarrow \frac{a}{5a+b} = \frac{c}{5c+d} \Longrightarrow ad=bc \Longrightarrow \frac{a}{b} = \frac{c}{d} =k \Longrightarrow a=bk \wedge c=dk \Longrightarrow \frac{7a+b}{7c+d} =8 \Longrightarrow \frac{b(7k+1)}{d(7k+1)} =8 \Longrightarrow \frac{b}{d} =8 \Longrightarrow \frac{9a+b}{9c+d} = \frac{9bk+b}{9dk+d} = \frac{b(9k+1)}{d(9k+1)}= \frac{b}{d} \Longrightarrow ?=8$$