726° Вправа 726Алгебра 8 клас Тарасенкова 2025 · § 17 · стор. 215

726. Зведіть рівняння до вигляду ax² + bx + c = 0:

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

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

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

2) ${\displaystyle 8x=}$${\displaystyle (x-4)(x+4)}$;

${\displaystyle 8x=}$${\displaystyle x^2-16}$;

${\displaystyle -x^2+8x+16=}$${\displaystyle 0}$, або ${\displaystyle x^2-8x-16=}$${\displaystyle 0}$;

3) ${\displaystyle {(x+2)}^2-13=}$${\displaystyle 3(x-1)}$;

${\displaystyle x^2+4x+4-13=}$${\displaystyle 3x-3}$;

${\displaystyle x^2+4x-9-3x+3=}$${\displaystyle 0}$;

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

4) ${\displaystyle 4x^2-5=}$${\displaystyle 2x(1+3x)}$;

${\displaystyle 4x^2-5=}$${\displaystyle 2x+6x^2}$;

${\displaystyle 4x^2-5-2x-6x^2=}$${\displaystyle 0}$;

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

5) ${\displaystyle {(x+3)}^2-6=}$${\displaystyle 8+10x}$;

${\displaystyle x^2+6x+9-6-8-10x=}$${\displaystyle 0}$;

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

6) ${\displaystyle 4(x+8)=}$${\displaystyle (x-4)(x+2)}$;

${\displaystyle 4x+32=}$${\displaystyle x^2+2x-4x-8}$;

${\displaystyle 4x+32-x^2+2x+8=}$${\displaystyle 0}$;

${\displaystyle -x^2+6x+40=}$${\displaystyle 0}$, або ${\displaystyle x^2-6x-40=}$${\displaystyle 0}$.