395 Вправа 395Алгебра 8 клас Бевз 2025 · § 8 · стор. 80

395. Розв’яжи рівняння.

а) ${\displaystyle \frac{5}{2-2t}-\frac{5}{18}=}$${\displaystyle \frac{8}{3t-3}-\frac{2+t}{t-1}}$;

${\displaystyle \frac{5}{2-2t}-\frac{5}{18}-\frac{8}{3t-3}+\frac{2+t}{t-1}=}$${\displaystyle 0}$;

${\displaystyle -\frac{5^{\mathrm{\setminus }9}}{2(t-1)}-\frac{5^{\mathrm{\setminus }t-1}}{18}-\frac{8^{\mathrm{\setminus }6}}{3(t-1)}+\frac{{2+t}^{\mathrm{\setminus }18}}{t-1}}$=

=${\displaystyle \frac{-45-5(t-1)-48+18(2+t)}{18(t-1)}}$=

=${\displaystyle \frac{-45-5t+5-48+18t+36}{18(t-1)}=}$${\displaystyle \frac{13t-52}{18(t-1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 13t-52=0\\ & 18(t-1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& t=4\\ & t\ne 1\end{array}}$;

${\displaystyle t=}$${\displaystyle 4}$;

б) ${\displaystyle \frac{6n+5}{4n+3}+\frac{3n-7}{3-4n}+\frac{6-4n^2+30n}{9-16n^2}=}$${\displaystyle 0}$;

${\displaystyle \frac{6n+5}{4n+3}+\frac{3n-7}{3-4n}+\frac{6-4n^2+30n}{9-16n^2}=}$${\displaystyle 0}$;

${\displaystyle \frac{{6n+5}^{\mathrm{\setminus }3-4n}}{4n+3}+\frac{{3n-7}^{\mathrm{\setminus }4n+3}}{3-4n}+\frac{6-4n^2+30n}{(3-4n)(4n+3)}}$=

=${\displaystyle \frac{(6n+5)(3-4n)+(3n-7)(4n+3)+6-4n^2+30n}{(3-4n)(4n+3)}}$=

=${\displaystyle \frac{15-24n^2-2n+12n^2-19n-21+30n-4n^2+6}{(3-4n)(4n+3)}}$=

=${\displaystyle \frac{9n-16n^2}{(3-4n)(4n+3)}}$=0;

${\displaystyle \{\begin{array}{rl}& -n(16n-9)=0\\ & (3-4n)(4n+3)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& n=0\text{або }n=\frac{9}{16}\\ & n\ne -\frac{3}{4},n\ne \frac{3}{4}\end{array}}$;

${\displaystyle n=}$${\displaystyle 0\text{або }n=}$${\displaystyle \frac{9}{16}}$.