174 Вправа 174Алгебра 8 клас Бевз 2025 · § 4 · стор. 38

174. Розв’яжи рівняння.

а) ${\displaystyle \frac{{3x-6}^{\mathrm{\setminus }2}}{3x}-\frac{{2-x}^{\mathrm{\setminus }3}}{2x}=}$${\displaystyle 0}$;

${\displaystyle \frac{2(3x-6)-3(2-x)}{6x}=}$${\displaystyle \frac{6x-12-6+3x}{6x}=}$${\displaystyle \frac{9x-18}{6x}}$=0;

${\displaystyle \{\begin{array}{rl}& 9x-18=0\\ & 6x\ne 0\end{array}}$;

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

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

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

${\displaystyle \frac{x^{\mathrm{\setminus }2}}{3(x-1)}+\frac{5^{\mathrm{\setminus }3}}{2(x-1)}=}$${\displaystyle \frac{2x+15}{6(x-1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 2x+15=0\\ & 6(x-1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{15}{2}\\ & x\ne 1\end{array}}$;

${\displaystyle x=}$${\displaystyle -7,5}$;

в) ${\displaystyle \frac{x}{{(x+3)}^2}+\frac{2^{\mathrm{\setminus }x+3}}{x+3}=}$${\displaystyle 0}$;

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

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

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

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

г) ${\displaystyle \frac{10+x^2}{x-x^2}+\frac{x+5}{x-1}=}$${\displaystyle 0}$;

${\displaystyle -\frac{10+x^2}{x(x-1)}+\frac{{x+5}^{\mathrm{\setminus }x}}{x-1}=}$${\displaystyle \frac{-(10+x^2)+x(x+5)}{x(x-1)}}$=

=${\displaystyle \frac{-x^2-10+x^2+5x}{x(x-1)}=}$${\displaystyle \frac{5x-10}{x(x-1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 5x-10=0\\ & x(x-1)\ne 0\end{array}}$;

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

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