837° Вправа 837Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 253

837. Розв’яжіть рівняння:

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

${\displaystyle 2x^2-10x+5-x=}$${\displaystyle 0}$;

${\displaystyle 2x^2-11x+5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-11)}^2-4\cdot 2\cdot 5=}$${\displaystyle 121-40=}$${\displaystyle 81}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{11\pm \sqrt{81}}{2\cdot 2}=}$${\displaystyle \frac{11\pm 9}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{11+9}{4}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-9}{4}=}$${\displaystyle \frac{2}{4}=}$${\displaystyle 0,5}$;

2) ${\displaystyle {(2x-1)}^2+5x=}$${\displaystyle 3x^2+7}$;

${\displaystyle 4x^2-4x+1+5x-3x^2-7=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-6)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

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

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

${\displaystyle x_2=}$${\displaystyle \frac{-1-5}{2}=}$${\displaystyle -3}$.

3) ${\displaystyle {(x-2)}^2+{(x+2)}^2=}$${\displaystyle 3x+7}$;

${\displaystyle x^2-4x+4+x^2+4x+4-3x-7=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 2\cdot 1=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{3\pm \sqrt{1}}{2\cdot 2}=}$${\displaystyle \frac{3\pm 1}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+1}{4}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-1}{4}=}$${\displaystyle \frac{2}{4}=}$${\displaystyle 0,5}$;

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

${\displaystyle 4x^2-12x+9-4=}$${\displaystyle x^2-4x}$;

${\displaystyle 4x^2-x^2-12x+4x+5=}$${\displaystyle 0}$;

${\displaystyle 3x^2-8x+5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-8)}^2-4\cdot 3\cdot 5=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{8\pm \sqrt{4}}{2\cdot 3}=}$${\displaystyle \frac{8\pm 2}{6}}$;

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

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