Toán 11

K

khongphaibang

ta co : /sinx/\leq1\Rightarrowsin^2008x\leqsin^2x(1)

/cosx\leq1\Rightarrowcosx\leq/cosx/\Rightarrowcos^2009x\leq/cosx/^2009(2)

cong hai ve cua (1)va (2) ta duoc :

sin^2008x + cos^2009x\leqsin^2x+cos^2x (/cosx/^2=cos^2x)
Dau bang xay ra \Leftrightarrow sin^2008x = sin^2x(3) va cos^2009x=cos^2x(4)

giai (3): sin^2008x=sin^2x
\Leftrightarrowsin^2008x-sin^2x=0
\Leftrightarrowsin^2x.(sin^2006x-1)=0
\Rightarrowsin^2x=0 hoac sin^2006x=1

Ban tu giai ra ket qua va thu lai vao (4) xem co thoa man ko roi ket luon ngiem)
 
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