134 Вправа 134Алгебра 8 клас Бевз 2025 · § 3 · стор. 30

134. Розв’яжи рівняння.

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

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

${\displaystyle \{\begin{array}{rl}& x=0\\ & x\ne \pm 1\end{array}}$;

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

б) ${\displaystyle \frac{x^2-1}{x+7}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& (x-1)(x+1)=0\\ & x+7\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=\pm 1\\ & x\ne -7\end{array}}$;

${\displaystyle x=}$${\displaystyle \pm 1}$;

в) ${\displaystyle \frac{x^2-3x}{x^2-6x+9}=}$${\displaystyle 0}$;

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

${\displaystyle \{\begin{array}{rl}& x=0\text{або }x=3\\ & x\ne 3\end{array}}$;

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

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

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

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

коренів немає.