Question

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Answer 1

Ответ:   x=2 .

$2x^2-8x+7<0\\\\D=8\ \ ,\ \ x_1=\dfrac{8-2\sqrt2}{4}=2-\dfrac{\sqrt2}{2}\approx 1,3\ \ ,\ \ x_2=2+\dfrac{\sqrt2}{2}\approx 2,7\\\\x\in \Big(\ 2-\dfrac{\sqrt2}{2}\ ;\ 2+\dfrac{\sqrt2}{2}\ \Big)$

Целые решения неравенства:  $x=2$  .