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