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thienluan14211

$x+y=1$ \Rightarrow Chỗ nào có x+y thì cứ thay bằng 1
$y^3+y^3=a$
$x^5+y^5=b$

Ta có
$a=x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)^2-3xy=1-3xy$
$b=x^5+y^5=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)$
$=(x^4+y^4)-xy(x^2+y^2)+x^2y^2$
$=(x^2+y^2)^2-2x^2y^2-xy(x^2+y^2)+x^2y^2$
$=$[$(x+y)^2-2xy$]$ ^2-2x^2y^2-xy[(x+y)^2-2xy$]$+x^2y^2$
$=(1-2xy)^2-xy(1-2xy)-x^2y^2$
$=1+5x^2y^2-5xy$

$5a(a+1)=5(1-3x)(2-3x)=10-45xy+45x^2y^2$
$9b+1=9+45x^2y^2-45xy+1=10-45xy+45x^2y^2$
\Rightarrow5a(a+1)=9b+1
 
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