122 Вправа 122Алгебра 8 клас Бевз 2025 · § 3 · стор. 29

122. Розв’яжи рівняння.

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

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

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

${\displaystyle x=}$${\displaystyle -3}$;

б) ${\displaystyle \frac{2x-10}{x+5}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& 2x-10=0\\ & x+5\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=5\\ & x\ne -5\end{array}}$;

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

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

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

${\displaystyle \{\begin{array}{rl}& x=0\text{або }x=-1\\ & x\ne 1,x\ne -1\end{array}}$;

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

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

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