160 Вправа 160Алгебра 8 клас Бевз 2025 · § 4 · стор. 36

160. Розв’яжи рівняння.

а) ${\displaystyle \frac{8}{3+x}+\frac{2x}{3+x}=}$${\displaystyle 0}$;

${\displaystyle \frac{8}{x+3}+\frac{2x}{x+3}=}$${\displaystyle \frac{8+2x}{x+3}}$=0;

${\displaystyle \{\begin{array}{rl}& 2x+8=0\\ & x+3\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-4\\ & x\ne -3\end{array}}$;

${\displaystyle x=}$${\displaystyle -4}$;

б) ${\displaystyle \frac{2x}{x-2}-\frac{4}{x-2}=}$${\displaystyle 0}$;

${\displaystyle \frac{2x-4}{x-2}}$=0;

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

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

коренів немає;

в) ${\displaystyle \frac{5x-1}{2x+5}+\frac{2x+15}{2x+5}=}$${\displaystyle 0}$;

${\displaystyle \frac{5x-1+2x+15}{2x+5}=}$${\displaystyle \frac{7x+14}{2x+5}}$=0;

${\displaystyle \{\begin{array}{rl}& 7x+14=0\\ & 2x+5\ne 0\end{array}}$;

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

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