396 Вправа 396Алгебра 8 клас Бевз 2025 · § 8 · стор. 80

396. Розв’яжи рівняння.

а) ${\displaystyle \frac{3}{8-2x}-\frac{5}{6}=}$${\displaystyle \frac{14}{3x-12}-\frac{2+x}{x-4}}$;

${\displaystyle \frac{3}{8-2x}-\frac{5}{6}-\frac{14}{3x-12}+\frac{2+x}{x-4}=}$${\displaystyle 0}$;

${\displaystyle -\frac{3^{\mathrm{\setminus }3}}{2(x-4)}-\frac{5^{\mathrm{\setminus }x-4}}{6}-\frac{{14}^{\mathrm{\setminus }2}}{3(x-4)}+\frac{{2+x}^{\mathrm{\setminus }6}}{x-4}}$=

=${\displaystyle \frac{-9-5(x-4)-28+6(2+x)}{6(x-4)}}$=

=${\displaystyle \frac{-9-5x+20-28+6x+12}{6(x-4)}=}$${\displaystyle \frac{x-5}{6(x-4)}}$=0;

${\displaystyle \{\begin{array}{rl}& x-5=0\\ & 6(x-4)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=5\\ & x\ne 4\end{array}}$;

${\displaystyle x=}$${\displaystyle 5}$;

б) ${\displaystyle \frac{2z-1}{z+4}-\frac{3z-1}{4-z}-\frac{96}{z^2-16}=}$${\displaystyle 5}$;

${\displaystyle \frac{2z-1}{z+4}-\frac{3z-1}{4-z}-\frac{96}{z^2-16}-5=}$${\displaystyle 0}$;

${\displaystyle \frac{{2z-1}^{\mathrm{\setminus }z-4}}{z+4}+\frac{{3z-1}^{\mathrm{\setminus }z+4}}{z-4}-\frac{96}{(z-4)(z+4)}-\frac{5^{\mathrm{\setminus }\left(z-4\right)\left(z+4\right)}}{1}}$=

=${\displaystyle \frac{(2z-1)(z-4)+(3z-1)(z+4)-96-5(z-4)(z+4)}{(z-4)(z+4)}}$=

=${\displaystyle \frac{2z^2-9z+4+3z^2+11z-4-96-5z^2+80}{(z-4)(z+4)}}$=

=${\displaystyle \frac{2z-16}{(z-4)(z+4)}}$=0;

${\displaystyle \{\begin{array}{rl}& 2z-16=0\\ & (z-4)(z+4)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& z=8\\ & z\ne -4,z\ne 4\end{array}}$;

${\displaystyle z=}$${\displaystyle 8}$.