843° Вправа 843Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 254

843. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x^2+2x-15}{x^2+x-12}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& x^2+2x-15=0,\\ & x^2+x-12\ne 0\end{array}}$;

${\displaystyle x^2+2x-15=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle -2}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -15}$;

${\displaystyle x_1=}$${\displaystyle -5}$; ${\displaystyle x_2=}$${\displaystyle 3}$;

${\displaystyle x^2+x-12\ne 0}$;

${\displaystyle (x+4)(x-3)\ne 0}$;

${\displaystyle x+4\ne 0}$ і ${\displaystyle x-3\ne 0}$;

${\displaystyle x\ne -4}$, ${\displaystyle x\ne 3}$;

${\displaystyle x=}$${\displaystyle -5}$;

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

${\displaystyle \{\begin{array}{rl}& 2x^2+3x-2=0,\\ & x^2+2x\ne 0\end{array}}$;

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

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

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-3\pm \sqrt{25}}{2\cdot 2}=}$${\displaystyle \frac{-3\pm 5}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+5}{4}=}$${\displaystyle \frac{2}{4}=}$${\displaystyle 0,5}$;

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

${\displaystyle x^2+2x\ne 0}$;

${\displaystyle x(x+2)\ne 0}$;

${\displaystyle x\ne 0}$ і ${\displaystyle x+2\ne 0}$;

${\displaystyle x\ne 0}$, ${\displaystyle x\ne -2}$;

${\displaystyle x=}$${\displaystyle 0,5}$.