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$x^5+y^5=(x+y)(x^4-x^3y-xy^3+x^2y^2+y^4)=[(x^2+y^2)^2-x^2y^2-xy(x^2+y^2)]=(x+y)(1-x^2y^2-xy)$
$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$
VT=$|16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|$
=$|(x+y)(16-16x^2y^2-16xy-20+20xy+5)|$
=$|(x+y)(-16x^2y^2+4xy+1)|$ (1)
Theo gt $x^2+y^2=1\Leftrightarrow (x+y)^2-2xy=1\Rightarrow \frac{(x+y)^2-1}{2}=xy$
Đặt t=x+y ($-\sqrt{2}\leq t\leq \sqrt{2}$)
(1)=$|t(-16(\frac{t^2-1}{2})^2+4\frac{t^2-1}{2}+1)|$
=$|-4t^5+10t^3-5t|$
Xét $ f(t)=-4t^5+10t^3-5t$ ($-\sqrt{2}\leq t\leq \sqrt{2}$)
$f'(t)=-20t^4+30t^2-5=-5[(2t^2-\frac{3}{2})^2+\frac{11}{4}]<0 \forall t$
$\Rightarrow $hs nghịch biến
$t -\sqrt{2} 0 \sqrt{2} $
$f'(t) | - |$
$. |\sqrt{2} |$
$. | 0 |$
$f(t) | -\sqrt{2} |$
$f(t)\in [-\sqrt{2} ;\sqrt{2}]\Rightarrow |(x+y)(-16x^2y^2+4xy+1)|\in [0;\sqrt{2}]$
$\Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)
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