838° Вправа 838Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 253

838. Знайдіть усі значення змінної х, за яких:

1) ${\displaystyle (x-5)(x+5)+4=}$${\displaystyle 14(x-5)}$;

${\displaystyle x^2-25+4=}$${\displaystyle 14x-70}$;

${\displaystyle x^2-21-14x+70=}$${\displaystyle 0}$;

${\displaystyle x^2-14x+49=}$${\displaystyle 0}$;

${\displaystyle {(x-7)}^2=}$${\displaystyle 0}$;

${\displaystyle x-7=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 7}$;

2) ${\displaystyle {(x+4)}^2+(x+6)=}$${\displaystyle 22}$;

${\displaystyle x^2+8x+16+x+6-22=}$${\displaystyle 0}$;

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

${\displaystyle x(x+9)=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x+9=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle 0}$; ${\displaystyle x_2=}$${\displaystyle -9}$.

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

${\displaystyle 6-2x-1=}$${\displaystyle 3x^2-4x}$;

${\displaystyle 3x^2-4x+2x-5=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 3\cdot (-5)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{2\pm \sqrt{64}}{2\cdot 3}=}$${\displaystyle \frac{2\pm 8}{6}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{2-8}{6}=}$${\displaystyle -1}$;

4) ${\displaystyle 3{(x+2)}^2-{(x+3)}^2=}$${\displaystyle 11}$;

${\displaystyle 3(x^2+4x+4)-(x^2+6x+9)-11=}$${\displaystyle 0}$;

${\displaystyle 3x^2+12x+12-x^2-6x-9-11=}$${\displaystyle 0}$;

${\displaystyle 2x^2+6x-8=}$${\displaystyle 0}$ | ${\displaystyle :2}$

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

${\displaystyle D=}$${\displaystyle 3^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-3\pm \sqrt{25}}{2\cdot 1}=}$${\displaystyle \frac{-3\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+5}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-5}{2}=}$${\displaystyle -4}$.