(4x^2 +2x+1)^(x^2 −x) >1


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Question Number 198228 by sulaymonnorboyev140 last updated on 14/Oct/23

${(4 x^{2} + 2 x + 1)}^{x^{2} - x} > 1$
	Answered by MM42 last updated on 14/Oct/23
	${ ((4x^2 +2x+1>1)),((x^2 −x>0)) :}⇒(−∞,−(1/2))∪(0,+∞)=A & (−∞,0)∪(1,+∞)=B ⇒A∩B=(−∞,−(1/2))∪(1,+∞) (i) { ((0<4x^2 +2x+1<1)),((x^2 −x<0)) :}⇒(−(1/2),0)=C & (0,1)=D ⇒C∩D=φ (ii) (i)∪(ii)=(i)$
	${\begin{cases} 4 x^{2} + 2 x + 1 > 1 \\ x^{2} - x > 0 \end{cases} \Rightarrow (- \infty, - \frac{1}{2}) \cup (0, + \infty) = A & (- \infty, 0) \cup (1, + \infty) = B$ $\Rightarrow A \cap B = (- \infty, - \frac{1}{2}) \cup (1, + \infty) (i)$ ${\begin{cases} 0 < 4 x^{2} + 2 x + 1 < 1 \\ x^{2} - x < 0 \end{cases} \Rightarrow (- \frac{1}{2}, 0) = C & (0, 1) = D$ $\Rightarrow C \cap D = ϕ (i i)$ $(i) \cup (i i) = (i)$
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