819 Вправа 819Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 244

819. Скоротіть дріб:

1)

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot (-6)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{49}}{2\cdot 1}=}$${\displaystyle \frac{5\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+7}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-7}{2}=}$${\displaystyle -1}$;

${\displaystyle x^2-5x-6=}$${\displaystyle (x+1)(x-6)}$;

${\displaystyle \frac{4x+4}{x^2-5x-6}=}$${\displaystyle \frac{4\cancel{\left(x+1\right)}}{\cancel{\left(x+1\right)}(x-6)}=}$${\displaystyle \frac{4}{x-6}}$;

2)

${\displaystyle 3x^2-10x+3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-10)}^2-4\cdot 3\cdot 3=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{10\pm \sqrt{64}}{2\cdot 3}=}$${\displaystyle \frac{10\pm 8}{6}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{10-8}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle 3x^2-10x+3=}$${\displaystyle 3(x-3)(x-\frac{1}{3})=}$${\displaystyle (x-3)(3x-1)}$;

${\displaystyle \frac{x^2-3x}{3x^2-10x+3}=}$${\displaystyle \frac{x\cancel{\left(x-3\right)}}{\cancel{\left(x-3\right)}(3x-1)}=}$${\displaystyle \frac{x}{3x-1}}$;

3)

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

${\displaystyle D=}$${\displaystyle {(-4)}^2-4\cdot 5\cdot (-12)=}$${\displaystyle 16+240=}$${\displaystyle 256}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{4\pm \sqrt{256}}{2\cdot 5}=}$${\displaystyle \frac{4\pm 16}{10}}$;

${\displaystyle x_1=}$${\displaystyle \frac{4+16}{10}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{4-16}{10}=}$${\displaystyle \frac{-12}{10}=}$${\displaystyle -1,2}$;

${\displaystyle 5x^2-4x-12=}$${\displaystyle 5(x-2)(x+1,2)=}$${\displaystyle (x-2)(5x+6)}$;

${\displaystyle \frac{5x^2-4x-12}{x^2-4}=}$${\displaystyle \frac{\cancel{\left(x-2\right)}(5x+6)}{\cancel{\left(x-2\right)}(x+2)}=}$${\displaystyle \frac{5x+6}{x+2}}$.