204 Вправа 204Алгебра 8 клас Бевз 2025 · § 4 · стор. 41

204. Розв’яжи рівняння.

а) ${\displaystyle \frac{6}{x^2-6x}+\frac{12+x}{6x-36}=}$${\displaystyle 0}$;

${\displaystyle \frac{6^{\mathrm{\setminus }6}}{x(x-6)}+\frac{{12+x}^{\mathrm{\setminus }x}}{6(x-6)}=}$${\displaystyle \frac{36+x(12+x)}{6x(x-6)}}$=

=${\displaystyle \frac{36+x^2+12x}{6x(x-6)}=}$${\displaystyle \frac{x^2+12x+36}{6x(x-6)}}$=0;

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

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

${\displaystyle x=}$${\displaystyle -6}$;

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

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

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

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

${\displaystyle \{\begin{array}{rl}& x=\frac{1}{5}\\ & x\ne -1,x\ne 0,x\ne 1\end{array}}$;

${\displaystyle x=}$${\displaystyle 0,2}$;

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

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

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

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

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

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

г) ${\displaystyle \frac{x^2-8}{x^3+8}-\frac{1}{x+2}=}$${\displaystyle 0}$;

${\displaystyle \frac{x^2-8}{(x+2)(x^2-2x+4)}-\frac{1^{\mathrm{\setminus }{x}^2-2x+4}}{x+2}=}$${\displaystyle \frac{x^2-8-(x^2-2x+4)}{(x+2)(x^2-2x+4)}}$=

=${\displaystyle \frac{x^2-8-x^2+2x-4}{(x+2)(x^2-2x+4)}=}$${\displaystyle \frac{2x-12}{(x+2)(x^2-2x+4)}}$=0;

${\displaystyle \{\begin{array}{rl}& 2x-12=0\\ & (x+2)(x^2-2x+4)\ne 0\end{array}}$;

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

${\displaystyle x=}$${\displaystyle 6}$.