Em làm cách này được không ạ?!
Với \(x\ne\pm y\), ta có: \(\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4+y^4}+\frac{8y^8}{x^8-y^8}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4\left(x^4-y^4\right)+8y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^2\left(x^4+y^4\right)}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4-y^4}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2\left(x^2-y^2\right)+4y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2\left(x^2+y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2-y^2}=4\)
\(\Leftrightarrow\frac{y\left(x-y\right)+2y^2}{\left(x-y\right)\left(x+y\right)}=4\)
\(\Leftrightarrow\frac{y\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=4\)
\(\Leftrightarrow\frac{y}{x-y}=4\)
\(\Leftrightarrow y=4x-4y\Leftrightarrow5y=4x\left(đpcm\right)\)
