Question

I
시
2
$- x^2 > \frac{1}{2} + 2x$
​

Answer 1

Resposta:

S = { x ∈ IR/ 1 - √2/2 < x < 1 + √2/2}

Explicação passo-a-passo:

$-x^2>\frac{1}{2} +2x\\\\-x^2-2x-\frac{1}{2}>0~~~~ * (-1)\\\\x^2+2x+\frac{1}{2}<0\\\\2x^2+4x+1<0\\\\2x^2+4x+1=0\\\\\Delta=4^2-4.2.1\\\\\Delta=16-8\\\\D=8\\\\x=\frac{-4-2\sqrt{2} }{2.2} \\\\x=\frac{-4}{4} -\frac{2\sqrt{2} }{4}\\\\x=-1-\frac{\sqrt{2} }{2} \\\\ou\\\\x=-1+\frac{\sqrt{2} }{2}$

....................-1-√2/2____________1+√2/2...................

             +                            -                                 +

S = { x ∈ IR/ 1 - √2/2 < x < 1 + √2/2}