Question

Даю 50 балов. Помогите, пж

Answer 1

$\displaystyle\bf 2x^{2} -4x+1<0\\\\ 2x^{2} -4x+1=0\\\\ D=(-4)^{2}-4\cdot 2\cdot 1=16-8=8=(2\sqrt{2} )^{2} \\\\\\ x_{1} =\frac{4+2\sqrt{2} }{4} =1+0,5\sqrt{2} \\\\\\ x_{1} =\frac{4+2\sqrt{2} }{4} =1-0,5\sqrt{2} \\\\\\ 2x^{2} -4x+1=2\Big[x-(1+0,5\sqrt{2} )\Big]\Big[x-(1-0,5\sqrt{2} )\Big]\\\\\\ \Big[x-(1+0,5\sqrt{2} )\Big]\Big[x-(1-0,5\sqrt{2} )\Big]<0$

+ + + + + (1 - 0,5√2) - - - - - (1 + 0,5√2) + + + + +

                              ///////////

$\displaystyle\bf x\in(1-0,5\sqrt{2} \ ; 1+0,5\sqrt{2} )\\\\ 1-0,5\sqrt{2} \approx0,3\\\\ 1+0,5\sqrt{2} \approx1,7\\\\ Otvet:B \ 1$