839° Вправа 839Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 253

839. Знайдіть усі значення змінної х, за яких:

1) ${\displaystyle 8(x+3)+1=}$${\displaystyle {(x+2)}^2}$;

${\displaystyle 8x+24+1=}$${\displaystyle x^2+4x+4}$;

${\displaystyle x^2+4x+4-8x-25=}$${\displaystyle 0}$;

${\displaystyle x^2-4x-21=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-4)}^2-4\cdot 1\cdot (-21)=}$${\displaystyle 16+84=}$${\displaystyle 100}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{4\pm \sqrt{100}}{2\cdot 1}=}$${\displaystyle \frac{4\pm 10}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{4+10}{2}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{4-10}{2}=}$${\displaystyle -3}$;

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

${\displaystyle 3(x^2+2x+1)=}$${\displaystyle 6-2x}$;

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

${\displaystyle 3x^2+8x-3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 8^2-4\cdot 3\cdot (-3)=}$${\displaystyle 64+36=}$${\displaystyle 100}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-8\pm \sqrt{100}}{2\cdot 3}=}$${\displaystyle \frac{-8\pm 10}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-8+10}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-8-10}{6}=}$${\displaystyle -3}$.