GTLN

Question

Cho

$x^{2}+y^{2}=1$ . CM

$\left| {16(x^{5}+y^{5}) - 20(x^{3}+y^{3})+5(x+y)} \right|\leq \sqrt{2}$

gunbk92 · Answer 1 · 3/11/2013 11:22:49 AM

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

$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)[(x^2+y^2)-xy)=(x+y)(1-xy)$
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)

1 Đáp án