842° Вправа 842Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 254

842. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x^2+5x-6}{x^2+2x-3}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& x^2+5x-6=0,\\ & x^2+2x-3\ne 0\end{array}}$;

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

${\displaystyle x_1+x_2=}$${\displaystyle -5}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -6}$;

${\displaystyle x_1=}$${\displaystyle -6}$; ${\displaystyle x_2=}$${\displaystyle 1}$;

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

${\displaystyle (x-1)(x+3)\ne 0}$;

${\displaystyle x-1\ne 0}$ і ${\displaystyle x+3\ne 0}$;

${\displaystyle x\ne 1}$, ${\displaystyle x\ne -3}$;

${\displaystyle x=}$${\displaystyle -6}$;

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

${\displaystyle \{\begin{array}{rl}& x^2-3x+2=0,\\ & x^2-4x+3\ne 0\end{array}}$;

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

${\displaystyle x_1+x_2=}$${\displaystyle 3}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 2}$;

${\displaystyle x_1=}$${\displaystyle 1}$; ${\displaystyle x_2=}$${\displaystyle 2}$;

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

${\displaystyle (x-1)(x-3)\ne 0}$;

${\displaystyle x-1\ne 0}$ і ${\displaystyle x-3\ne 0}$;

${\displaystyle x\ne 1}$, ${\displaystyle x\ne 3}$;

${\displaystyle x=}$${\displaystyle 2}$.

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

${\displaystyle \{\begin{array}{rl}& 2x^2-5x+2=0,\\ & x^2-2x\ne 0\end{array}}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 2=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{9}}{2\cdot 2}=}$${\displaystyle \frac{5\pm 3}{4}}$;

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

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

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

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

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

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

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

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

${\displaystyle \{\begin{array}{rl}& 3x^2-5x+2=0,\\ & 1-x^2\ne 0\end{array}}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 3\cdot 2=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{1}}{2\cdot 3}=}$${\displaystyle \frac{5\pm 1}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+1}{6}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-1}{6}=}$${\displaystyle \frac{4}{6}=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle 1-x^2\ne 0}$;

${\displaystyle x^2\ne 1}$;

${\displaystyle x\ne 1}$, ${\displaystyle x\ne -1}$;

${\displaystyle x=}$${\displaystyle \frac{2}{3}}$.