Question

3решите уравнения: 2x^2 − 4√2x + 3 = 0x^2 − x(√7−2) − 2√7 = 02x^2 − 4x − 17 = 0

Answer 1

1) $2x^2-4\sqrt{2}x+3=0;\ \sqrt{2}x=t;\ t^2-4t+3=0;\ (t-3)(t-1)=0;\ \left [{{t=3} \atop {t=1}} \right. ;\ \left [ {{x=\frac{3}{\sqrt{2}}} \atop {x=\frac{1}{\sqrt{2}}}} \right.$

ответ: $\frac{3\sqrt{2}}{2};\ \frac{\sqrt{2}}{2}$

2) $x^2-x(\sqrt{7}-2)-2\sqrt{7}=0;\ (x-\sqrt{7})(x+2)=0;\ \left [ {{x=\sqrt{7}} \atop {x=-2}} \right.$

ответ: $\sqrt{7};\ -2$

3) $2x^2-4x-17=0;\ 4x^2-8x-34=0;\ ((2x)^2-2\cdot (2x)\cdot 2+2^2)-38=0;$

$(2x-2)^2=38;\ 2x-2=\pm\sqrt{38};\ 2x=2\pm\sqrt{38};\ x=\frac{2\pm\sqrt{38}}{2}$

ответ: $\frac{2\pm\sqrt{38}}{2}$