Theo đề ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\)\(\Rightarrow a\left(b+9\right)=b\left(a+6\right)\)
\(\Rightarrow ab+9a=ab+6b\)
\(\Rightarrow ab+9a-ab-6b=0\)
\(\Rightarrow9x-6y=0\)
\(\Rightarrow9x=6y\Rightarrow\frac{x}{y}=\frac{6}{9}=\frac{2}{3}\)
Vậy phân số đó là \(\frac{2}{3}\)
Theo đề ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\Rightarrow a\left(b+9\right)=b\left(a+6\right)\)
\(\Rightarrow ab+9a=ab+6b\)
\(\Rightarrow ab+9a-ab-6b=0\)
\(\Rightarrow9a-6b=0\)
\(\Rightarrow9a=6b\Rightarrow\frac{a}{b}=\frac{6}{9}=\frac{2}{3}\)
Vậy phân số phải tìm là \(\frac{2}{3}\)
Theo đề bài, ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\)
\(\frac{a\times\left(b+9\right)}{b\times\left(b+9\right)}=\frac{\left(a+6\right)\times b}{\left(b+9\right)\times b}\)
Vậy \(a\times\left(b+9\right)=\left(a+6\right)\times b\)
\(a\times b+a\times9=a\times b+6\times b\)
\(a\times9=6\times b\)
\(\frac{a\times9}{b\times9}=\frac{6\times b}{b\times9}\)
\(\frac{a}{b}=\frac{6}{9}\)
\(\frac{a}{b}=\frac{2}{3}\)
