206 Вправа 206Алгебра 8 клас Бевз 2025 · § 4 · стор. 42

206. Розв’яжи рівняння.

а) ${\displaystyle \frac{1}{x^2+3x}+\frac{3}{9-x^2}+\frac{1}{2x^2-6x}=}$${\displaystyle 0}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2\left(x-3\right)}}{x(x+3)}-\frac{3^{\mathrm{\setminus }2x}}{(x-3)(x+3)}+\frac{1^{\mathrm{\setminus }x+3}}{2x(x-3)}=}$${\displaystyle \frac{2(x-3)-6x+x+3}{2x(x-3)(x+3)}}$=

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

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

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

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

б) ${\displaystyle \frac{3x^2+7x+3}{x^3-1}-\frac{2x-1}{x^2+x+1}-\frac{1}{x-1}=}$${\displaystyle 0}$;

${\displaystyle \frac{3x^2+7x+3}{(x-1)(x^2+x+1)}-\frac{{2x-1}^{\mathrm{\setminus }x-1}}{x^2+x+1}-\frac{1^{\mathrm{\setminus }{x}^2+x+1}}{x-1}}$=

=${\displaystyle \frac{3x^2+7x+3-(2x-1)(x-1)-(x^2+x+1)}{(x-1)(x^2+x+1)}}$=

=${\displaystyle \frac{3x^2+7x+3-2x^2+3x-1-x^2-x-1}{(x-1)(x^2+x+1)}}$=

=${\displaystyle \frac{9x+1}{(x-1)(x^2+x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 9x+1=0\\ & (x-1)(x^2+x+1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{1}{9}\\ & x\ne 1\end{array}}$;

${\displaystyle x=}$${\displaystyle -\frac{1}{9}}$.