\(\frac{2a}{b}-\frac{2b}{a}=3\) (đề thế này à?)
\(\frac{2a}{b}-\frac{2b}{a}=3\)
\(\Leftrightarrow\frac{2a^2-2b^2}{ab}=3\)
\(\Leftrightarrow2a^2-2b^2=3ab\)
\(\Leftrightarrow a=\frac{2a^2-2b^2}{3b}\)
khi đó \(S=\frac{a-b}{a+b}=\frac{\frac{2a^2-2b^2}{3b}-b}{\frac{2a^2-2b^2}{3b}+b}=\frac{2a^2-2b^2-3b^2}{\frac{3b}{\frac{2a^2-2b^2+3b^2}{3b}}}=\frac{2a^2-5b^2}{3b}.\frac{3b}{2a^2+b^2}=\frac{2a^2-5b^2}{2a^2+b^2}\)
\(=\frac{2a^2+b^2-6b^2}{2a^2+b^2}=1-\frac{6b^2}{2a^2+b^2}\)
mk chịu....đề hơi kì