Question

3. Выделите полный квадрат: 2x2 - 3x +1

Answer 1

$2x^2-3x+1=(\sqrt2x)^2-2\cdot \sqrt2x\cdot \dfrac{3}{2\sqrt2}+\Big(\dfrac{3}{2\sqrt2}\Big)^2-\Big(\dfrac{3}{2\sqrt2}\Big)^2+1==\Big(\sqrt2x-\dfrac{3}{2\sqrt2}\Big)^2-\dfrac{9}{8}+1=\Big(\sqrt2x-\dfrac{3\sqrt2}{4}\Big)^2-\dfrac{1}{8}=\Big(\sqrt2\Big(x-\dfrac{3}{4}\Big)\Big)^2-\dfrac{1}{8}==2\Big(x-\dfrac{3}{4}\Big)^2-\dfrac{1}{8}$

или

  $\displaystyle 2x^2-3x+1=2\Big(x^2-\frac{3}{2}\, x\Big)+1=2\Big(\Big(x-\dfrac{3}{4}\Big)^2-\frac{9}{16}\Big)+1==2\Big(x-\dfrac{3}{4}\Big)^2-\frac{9}{8}+1=2\Big(x-\dfrac{3}{4}\Big)^2-\frac{1}{8}$