đại số 8

C

congchuaanhsang

Ta có: $x^5$+$y^5$=$2x^3y^3$\Leftrightarrow$\frac{x^5+y^5}{2x^3y^3}$=1
\Rightarrow$( \frac{x^5+y^5}{2x^3y^3} )^2$=1
Do đó H=1-$\frac{1}{xy}$=$( \frac{x^5+y^5}{2x^3y^3} )^2$-$\frac{1}{xy}$
=$\frac{(x^5+y^5)^2-4x^5y^5}{4x^6y^6}$
=$\frac{x^{10}+2x^5y^5+y^{10}-4x^5y^5}{4x^6y^6}$
=$\frac{x^{10}-2x^5y^5+y^{10}}{4x^6y^6}$
=$( \frac{x^5-y^5}{2x^3y^3} )^2$
 
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