Từ \(\frac{a+2}{a-2}=\frac{b+3}{b-3}\)
<=> (a+2)(b-3) = (a-2)(b+3)
<=> ab-3a+2b-6 = ab+3a-2b-6
<=> -6a = -4b
<=> \(\frac{a}{b}=\frac{3}{2}\)
<=> \(\frac{a}{2}=\frac{b}{3}\)
Ta có:\(\frac{a+2}{a-2}=\frac{b+3}{b-3}\)
\(\Leftrightarrow\frac{a-2+4}{a-2}=\frac{b-3+6}{b-3}\)
\(\Leftrightarrow1+\frac{4}{a-2}=1+\frac{6}{b-3}\)
\(\Leftrightarrow\frac{4}{a-2}=\frac{6}{b-3}\)
\(\Leftrightarrow\frac{2}{a-2}=\frac{3}{b-3}\)
\(\Leftrightarrow\frac{a-2}{2}=\frac{b-3}{3}\)
\(\Leftrightarrow\frac{a}{2}-1=\frac{b}{3}-1\)
\(\Leftrightarrow\frac{a}{2}=\frac{b}{3}\)
Giải:
Ta có: \(\frac{a+2}{a-2}=\frac{b+3}{b-3}\Rightarrow\frac{a+2}{b+3}=\frac{a-2}{b-3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+2}{b+3}=\frac{a-2}{b-3}=\frac{a}{b}=\frac{2}{3}\)
Vậy \(\frac{a}{b}=\frac{2}{3}\)