390 Вправа 390Алгебра 8 клас Бевз 2025 · § 8 · стор. 79

390. Знайди корені рівняння.

а) ${\displaystyle \frac{4}{1+x}-\frac{x+1}{x-1}=}$${\displaystyle \frac{x^2-3}{1-x^2}}$;

${\displaystyle \frac{4}{1+x}-\frac{x+1}{x-1}-\frac{x^2-3}{1-x^2}=}$${\displaystyle 0}$;

${\displaystyle \frac{4^{\mathrm{\setminus }x-1}}{x+1}-\frac{{x+1}^{\mathrm{\setminus }x+1}}{x-1}+\frac{x^2-3}{(x-1)(x+1)}}$=

=${\displaystyle \frac{4(x-1)-(x+1)(x+1)+x^2-3}{(x-1)(x+1)}}$=

=${\displaystyle \frac{4x-4-x^2-2x-1+x^2-3}{(x-1)(x+1)}=}$${\displaystyle \frac{2x-8}{(x-1)(x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 2x-8=0\\ & (x-1)(x+1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=4\\ & x\ne -1,x\ne 1\end{array}}$;

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

б) ${\displaystyle \frac{8+24x}{36x^2-1}+\frac{1}{1-6x}=}$${\displaystyle \frac{2}{6x+1}}$;

${\displaystyle \frac{8+24x}{36x^2-1}+\frac{1}{1-6x}-\frac{2}{6x+1}=}$${\displaystyle 0}$;

${\displaystyle \frac{8+24x}{(6x-1)(6x+1)}-\frac{1^{\mathrm{\setminus }6x+1}}{6x-1}-\frac{2^{\mathrm{\setminus }6x-1}}{6x+1}}$=

=${\displaystyle \frac{8+24x-(6x+1)-2(6x-1)}{(6x-1)(6x+1)}}$=

=${\displaystyle \frac{24x+8-6x-1-12x+2}{(6x-1)(6x+1)}=}$${\displaystyle \frac{6x+9}{(6x-1)(6x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 6x+9=0\\ & (6x-1)(6x+1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{3}{2}\\ & x\ne -\frac{1}{6},x\ne \frac{1}{6}\end{array}}$;

${\displaystyle x=}$${\displaystyle -1,5}$;

в) ${\displaystyle \frac{2}{1+x}-\frac{5}{x-1}=}$${\displaystyle \frac{7-x^2}{1-x^2}}$;

${\displaystyle \frac{2}{1+x}-\frac{5}{x-1}-\frac{7-x^2}{1-x^2}=}$${\displaystyle 0}$;

${\displaystyle \frac{2^{\mathrm{\setminus }x-1}}{x+1}-\frac{5^{\mathrm{\setminus }x+1}}{x-1}+\frac{7-x^2}{(x-1)(x+1)}}$=

=${\displaystyle \frac{2(x-1)-5(x+1)+7-x^2}{(x-1)(x+1)}}$=

=${\displaystyle \frac{2x-2-5x-5+7-x^2}{(x-1)(x+1)}=}$${\displaystyle \frac{-x^2-3x}{(x-1)(x+1)}}$=0;

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

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

${\displaystyle x=}$${\displaystyle -3\text{або }x=}$${\displaystyle 0}$;

г) ${\displaystyle \frac{3}{{(x+2)}^2}+\frac{2}{{(x-2)}^2}=}$${\displaystyle \frac{5}{x^2-4}}$;

${\displaystyle \frac{3}{{(x+2)}^2}+\frac{2}{{(x-2)}^2}-\frac{5}{x^2-4}=}$${\displaystyle 0}$;

${\displaystyle \frac{3^{\mathrm{\setminus }{\left(x-2\right)}^2}}{{(x+2)}^2}+\frac{2^{\mathrm{\setminus }{\left(x+2\right)}^2}}{{(x-2)}^2}-\frac{5^{\mathrm{\setminus }\left(x-2\right)\left(x+2\right)}}{(x-2)(x+2)}}$=

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

=${\displaystyle \frac{3x^2-12x+12+2x^2+8x+8-5x^2+20}{{(x-2)}^2{(x+2)}^2}}$=

=${\displaystyle \frac{40-4x}{{(x-2)}^2{(x+2)}^2}}$=0;

${\displaystyle \{\begin{array}{rl}& 40-4x=0\\ & {(x-2)}^2{(x+2)}^2\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=10\\ & x\ne -2,x\ne 2\end{array}}$;

${\displaystyle x=}$${\displaystyle 10}$.