394 Вправа 394Алгебра 8 клас Бевз 2025 · § 8 · стор. 79

394. Solve the equation 1/(1 − 3z)² − 3/(11 + 3z)² = 2/((1 − 3z)(11 + 3z)).

${\displaystyle \frac{1}{{(1-3z)}^2}-\frac{3}{{(11+3z)}^2}=}$${\displaystyle \frac{2}{(1-3z)(11+3z)}}$;

${\displaystyle \frac{1}{{(1-3z)}^2}-\frac{3}{{(11+3z)}^2}-\frac{2}{(1-3z)(11+3z)}=}$${\displaystyle 0}$;

${\displaystyle \frac{1^{\mathrm{\setminus }{\left(3z+11\right)}^2}}{{(3z-1)}^2}-\frac{3^{\mathrm{\setminus }{\left(3z-1\right)}^2}}{{(3z+11)}^2}+\frac{2^{\mathrm{\setminus }\left(3z-1\right)\left(3z+11\right)}}{(3z-1)(3z+11)}}$=

=${\displaystyle \frac{{(3z+11)}^2-3{(3z-1)}^2+2(3z-1)(3z+11)}{{(3z-1)}^2{(3z+11)}^2}}$=

=${\displaystyle \frac{9z^2+66z+121-27z^2+18z-3+18z^2+60z-22}{{(3z-1)}^2{(3z+11)}^2}}$=

=${\displaystyle \frac{144z+96}{{(3z-1)}^2{(3z+11)}^2}}$=0;

${\displaystyle \{\begin{array}{rl}& 144z+96=0\\ & {(3z-1)}^2{(3z+11)}^2\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& z=-\frac{2}{3}\\ & z\ne -\frac{11}{3},z\ne \frac{1}{3}\end{array}}$;

${\displaystyle z=}$${\displaystyle -\frac{2}{3}}$.