159 Вправа 159Алгебра 8 клас Бевз 2025 · § 4 · стор. 36

159. Розв’яжи рівняння.

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

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

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

${\displaystyle \{\begin{array}{rl}& x=3\\ & x\ne -5\end{array}}$;

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

б) ${\displaystyle \frac{2x}{x+3}+\frac{6}{x+3}=}$${\displaystyle 0}$;

${\displaystyle \frac{2x+6}{x+3}}$=0;

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

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

коренів немає;

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

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

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

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

коренів немає.