x5+y5=(x+y)(x4−x3y−xy3+x2y2+y4)=[(x2+y2)2−x2y2−xy(x2+y2)]=(x+y)(1−x2y2−xy)x3+y3=(x+y)(x2−xy+y2)=(x+y)[(x2+y2)−xy)=(x+y)(1−xy) VT=|16(x+y)(1−x2y2−xy)−20(x+y)(1−xy)+5(x+y)|=|(x+y)(16−16x2y2−16xy−20+20xy+5)|=|(x+y)(−16x2y2+4xy+1)| (1)Theo gt x2+y2=1⇔(x+y)2−2xy=1⇒(x+y)2−12=xy Đặt t=x+y (−√2≤t≤√2) (1)=|t(−16(t2−12)2+4t2−12+1)|=|−4t5+10t3−5t| Xét f(t)=−4t5+10t3−5t (−√2≤t≤√2) f′(t)=−20t4+30t2−5=−5[(2t2−32)2+114]<0∀t ⇒hs đồng biếnt−√20√2$f'(t) | + |$$. | \sqrt{2} |.|0|f(t) |-\sqrt{2} |f(t)∈[−√2;√2]⇒|(x+y)(−16x2y2+4xy+1)|∈[0;√2]\Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)
x5+y5=(x+y)(x4−x3y−xy3+x2y2+y4)=[(x2+y2)2−x2y2−xy(x2+y2)]=(x+y)(1−x2y2−xy)x3+y3=(x+y)(x2−xy+y2)=(x+y)[(x2+y2)−xy)=(x+y)(1−xy) VT=
|16(x+y)(1−x2y2−xy)−20(x+y)(1−xy)+5(x+y)|=
|(x+y)(16−16x2y2−16xy−20+20xy+5)|=
|(x+y)(−16x2y2+4xy+1)| (1)Theo gt
x2+y2=1⇔(x+y)2−2xy=1⇒(x+y)2−12=xy Đặt t=x+y (
−√2≤t≤√2) (1)=
|t(−16(t2−12)2+4t2−12+1)|=
|−4t5+10t3−5t| Xét
f(t)=−4t5+10t3−5t (
−√2≤t≤√2)
f′(t)=−20t4+30t2−5=−5[(2t2−32)2+114]<0∀t ⇒hs ng
hịch biến
t−√20√2$f'(t) |
- |$$. |
\sqrt{2} |
.|0|f(t) |
-\sqrt{2} |
f(t)∈[−√2;√2]⇒|(x+y)(−16x2y2+4xy+1)|∈[0;√2]\Rightarrow |16(x+y)( 1-x^2y^2-xy)-20(x+y)(1-xy)+5(x+y)|\leq \sqrt{2}$(đpcm)