[Toán 8]Chứng minh: $A =x^{12}-x^9+x^4-x+1>0$

N

nuocmat_nucuoi

TH1 : $x < 0 \Rightarrow -x^9 > 0 ; -x > 0$

$\Rightarrow x^{12}-x^9+x^4-x+1 > 0$

TH2 : $0 \leq x \leq 1$

$\Rightarrow x^4 \geq x^9 ; 1 \geq x$

$\Rightarrow x^{12}-x^9+x^4-x+1 \geq 0$

Dấu [=] không xảy ra nên $x^{12}-x^9+x^4-x+1 > 0$

TH3 : $ x > 1 \Rightarrow x^{12} > x^9 ; x^4 > x$

$\Rightarrow x^{12}-x^9+x^4-x+1 > 0$
 
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