734 Вправа 734Алгебра 8 клас Тарасенкова 2025 · § 17 · стор. 216

734. Розв’яжіть способом виділення квадрата двочлена рівняння:

1) ${\displaystyle x^2+3x-4=}$${\displaystyle 0}$;

${\displaystyle (x^2+2\cdot x\cdot 1,5+{1,5}^2)-2,25-4=}$${\displaystyle 0}$;

${\displaystyle {(x+1,5)}^2-6,25=}$${\displaystyle 0}$;

${\displaystyle {(x+1,5)}^2=}$${\displaystyle 6,25}$;

${\displaystyle x+1,5=}$${\displaystyle \sqrt{6,25}}$ або ${\displaystyle x+1,5=}$${\displaystyle -\sqrt{6,25}}$;

${\displaystyle x+1,5=}$${\displaystyle 2,5}$ або ${\displaystyle x+1,5=}$${\displaystyle -2,5}$;

${\displaystyle x=}$${\displaystyle 2,5-1,5}$ або ${\displaystyle x=}$${\displaystyle -2,5-1,5}$;

${\displaystyle x=}$${\displaystyle 1}$ або ${\displaystyle x=}$${\displaystyle -4}$;

2) ${\displaystyle x^2-5x+4=}$${\displaystyle 0}$;

${\displaystyle (x^2-2\cdot x\cdot 2,5+{2,5}^2)-6,25+4=}$${\displaystyle 0}$;

${\displaystyle {(x-2,5)}^2-2,25=}$${\displaystyle 0}$;

${\displaystyle {(x-2,5)}^2=}$${\displaystyle 2,25}$;

${\displaystyle x-2,5=}$${\displaystyle \sqrt{2,25}}$ або ${\displaystyle x-2,5=}$${\displaystyle -\sqrt{2,25}}$;

${\displaystyle x-2,5=}$${\displaystyle 1,5}$ або ${\displaystyle x-2,5=}$${\displaystyle -1,5}$;

${\displaystyle x=}$${\displaystyle 1,5+2,5}$ або ${\displaystyle x=}$${\displaystyle -1,5+2,5}$;

${\displaystyle x=}$${\displaystyle 4}$ або ${\displaystyle x=}$${\displaystyle 1}$;

3) ${\displaystyle x^2+x-6=}$${\displaystyle 0}$;

${\displaystyle (x^2+2\cdot x\cdot 0,5+{0,5}^2)-0,25-6=}$${\displaystyle 0}$;

${\displaystyle {(x+0,5)}^2-6,25=}$${\displaystyle 0}$;

${\displaystyle {(x+0,5)}^2=}$${\displaystyle 6,25}$;

${\displaystyle x+0,5=}$${\displaystyle \sqrt{6,25}}$ або ${\displaystyle x+0,5=}$${\displaystyle -\sqrt{6,25}}$;

${\displaystyle x+0,5=}$${\displaystyle 2,5}$ або ${\displaystyle x+0,5=}$${\displaystyle -2,5}$;

${\displaystyle x=}$${\displaystyle 2,5-0,5}$ або ${\displaystyle x=}$${\displaystyle -2,5-0,5}$;

${\displaystyle x=}$${\displaystyle 2}$ або ${\displaystyle x=}$${\displaystyle -3}$;

4) ${\displaystyle 2x^2-9x+10=}$${\displaystyle 0}$ | ${\displaystyle :2}$

${\displaystyle x^2-4,5x+5=}$${\displaystyle 0}$;

${\displaystyle (x^2-2\cdot x\cdot 2,25+{2,25}^2)-5,0625+5=}$${\displaystyle 0}$;

${\displaystyle {(x-2,25)}^2-0,0625=}$${\displaystyle 0}$;

${\displaystyle {(x-2,25)}^2=}$${\displaystyle 0,0625}$;

${\displaystyle x-2,25=}$${\displaystyle \sqrt{0,0625}}$ або ${\displaystyle x-2,25=}$${\displaystyle -\sqrt{0,0625}}$;

${\displaystyle x-2,25=}$${\displaystyle 0,25}$ або ${\displaystyle x-2,25=}$${\displaystyle -0,25}$;

${\displaystyle x=}$${\displaystyle 0,25+2,25}$ або ${\displaystyle x=}$${\displaystyle -0,25+2,25}$;

${\displaystyle x=}$${\displaystyle 2,5}$ або ${\displaystyle x=}$${\displaystyle 2}$;

5) ${\displaystyle 3x^2-5x+2=}$${\displaystyle 0}$ | ${\displaystyle :3}$

${\displaystyle x^2-\frac{5}{3}x+\frac{2}{3}=}$${\displaystyle 0}$;

${\displaystyle (x^2-2\cdot x\cdot \frac{5}{6}+{\left(\frac{5}{6}\right)}^2)-\frac{25}{36}+\frac{24}{36}=}$${\displaystyle 0}$;

${\displaystyle {(x-\frac{5}{6})}^2-\frac{1}{36}=}$${\displaystyle 0}$;

${\displaystyle {(x-\frac{5}{6})}^2=}$${\displaystyle \frac{1}{36}}$;

${\displaystyle x-\frac{5}{6}=}$${\displaystyle \sqrt{\frac{1}{36}}}$ або ${\displaystyle x-\frac{5}{6}=}$${\displaystyle -\sqrt{\frac{1}{36}}}$;

${\displaystyle x-\frac{5}{6}=}$${\displaystyle \frac{1}{6}}$ або ${\displaystyle x-\frac{5}{6}=}$${\displaystyle -\frac{1}{6}}$;

${\displaystyle x=}$${\displaystyle \frac{1}{6}+\frac{5}{6}}$ або ${\displaystyle x=}$${\displaystyle -\frac{1}{6}+\frac{5}{6}}$;

${\displaystyle x=}$${\displaystyle 1}$ або ${\displaystyle x=}$${\displaystyle \frac{2}{3}}$;

6) ${\displaystyle 5x^2+14x-3=}$${\displaystyle 0}$ | ${\displaystyle :5}$

${\displaystyle x^2+2,8x-0,6=}$${\displaystyle 0}$;

${\displaystyle (x^2+2\cdot x\cdot 1,4+{1,4}^2)-1,96-0,6=}$${\displaystyle 0}$;

${\displaystyle {(x+1,4)}^2-2,56=}$${\displaystyle 0}$;

${\displaystyle {(x+1,4)}^2=}$${\displaystyle 2,56}$;

${\displaystyle x+1,4=}$${\displaystyle \sqrt{2,56}}$ або ${\displaystyle x+1,4=}$${\displaystyle -\sqrt{2,56}}$;

${\displaystyle x+1,4=}$${\displaystyle 1,6}$ або ${\displaystyle x+1,4=}$${\displaystyle -1,6}$;

${\displaystyle x=}$${\displaystyle 1,6-1,4}$ або ${\displaystyle x=}$${\displaystyle -1,6-1,4}$;

${\displaystyle x=}$${\displaystyle 0,2}$ або ${\displaystyle x=}$${\displaystyle -3}$.