732 Вправа 732Алгебра 8 клас Тарасенкова 2025 · § 17 · стор. 216

732. Розв’яжіть способом виділення квадрата двочлена зведене квадратне рівняння:

1) ${\displaystyle x^2+2x-8=}$${\displaystyle 0}$;

${\displaystyle (x^2+2\cdot x\cdot 1+1^2)-1-8=}$${\displaystyle 0}$;

${\displaystyle {(x+1)}^2-9=}$${\displaystyle 0}$;

${\displaystyle {(x+1)}^2=}$${\displaystyle 9}$;

${\displaystyle x+1=}$${\displaystyle \sqrt{9}}$ або ${\displaystyle x+1=}$${\displaystyle -\sqrt{9}}$;

${\displaystyle x+1=}$${\displaystyle 3}$ або ${\displaystyle x+1=}$${\displaystyle -3}$;

${\displaystyle x=}$${\displaystyle 3-1}$ або ${\displaystyle x=}$${\displaystyle -3-1}$;

${\displaystyle x=}$${\displaystyle 2}$ або ${\displaystyle x=}$${\displaystyle -4}$;

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

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

${\displaystyle {(x-2)}^2-1=}$${\displaystyle 0}$;

${\displaystyle {(x-2)}^2=}$${\displaystyle 1}$;

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

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

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

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

3) ${\displaystyle x^2+12x+35=}$${\displaystyle 0}$;

${\displaystyle (x^2+2\cdot x\cdot 6+6^2)-36+35=}$${\displaystyle 0}$;

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

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

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

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

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

${\displaystyle x=}$${\displaystyle -5}$ або ${\displaystyle x=}$${\displaystyle -7}$;

4) ${\displaystyle x^2-8x+15=}$${\displaystyle 0}$;

${\displaystyle (x^2-2\cdot x\cdot 4+4^2)-16+15=}$${\displaystyle 0}$;

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

${\displaystyle {(x-4)}^2=}$${\displaystyle 1}$;

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

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

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

${\displaystyle x=}$${\displaystyle 5}$ або ${\displaystyle x=}$${\displaystyle 3}$;

5) ${\displaystyle x^2-6x-16=}$${\displaystyle 0}$;

${\displaystyle (x^2-2\cdot x\cdot 3+3^2)-9-16=}$${\displaystyle 0}$;

${\displaystyle {(x-3)}^2-25=}$${\displaystyle 0}$;

${\displaystyle {(x-3)}^2=}$${\displaystyle 25}$;

${\displaystyle x-3=}$${\displaystyle \sqrt{25}}$ або ${\displaystyle x-3=}$${\displaystyle -\sqrt{25}}$;

${\displaystyle x-3=}$${\displaystyle 5}$ або ${\displaystyle x-3=}$${\displaystyle -5}$;

${\displaystyle x=}$${\displaystyle 5+3}$ або ${\displaystyle x=}$${\displaystyle -5+3}$;

${\displaystyle x=}$${\displaystyle 8}$ або ${\displaystyle x=}$${\displaystyle -2}$;

6) ${\displaystyle x^2-10x+25=}$${\displaystyle 0}$;

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

${\displaystyle x-5=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 5}$.