388 Вправа 388Алгебра 8 клас Бевз 2025 · § 8 · стор. 79

388. Знайди корені рівняння.

а) ${\displaystyle \frac{12}{x^2-9}+\frac{2}{x+3}=}$${\displaystyle \frac{3}{x-3}}$;

${\displaystyle \frac{12}{x^2-9}+\frac{2}{x+3}-\frac{3}{x-3}=}$${\displaystyle 0}$;

${\displaystyle \frac{12}{(x-3)(x+3)}+\frac{2^{\mathrm{\setminus }x-3}}{x+3}-\frac{3^{\mathrm{\setminus }x+3}}{x-3}=}$${\displaystyle \frac{12+2(x-3)-3(x+3)}{(x-3)(x+3)}}$=

=${\displaystyle \frac{12+2x-6-3x-9}{(x-3)(x+3)}=}$${\displaystyle \frac{-x-3}{(x-3)(x+3)}}$=0;

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

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

коренів немає;

б) ${\displaystyle \frac{2x+4}{3-x}+1=}$${\displaystyle \frac{5-x}{x+2}}$;

${\displaystyle \frac{{2x+4}^{\mathrm{\setminus }x+2}}{3-x}+\frac{1^{\mathrm{\setminus }\left(3-x\right)\left(x+2\right)}}{1}-\frac{{5-x}^{\mathrm{\setminus }3-x}}{x+2}=}$${\displaystyle 0}$;

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

=${\displaystyle \frac{2x^2+8x+8+x-x^2+6-x^2+8x-15}{(3-x)(x+2)}=}$${\displaystyle \frac{17x-1}{(3-x)(x+2)}}$=0;

${\displaystyle \{\begin{array}{rl}& 17x-1=0\\ & (3-x)(x+2)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=\frac{1}{17}\\ & x\ne -2,x\ne 3\end{array}}$;

${\displaystyle x=}$${\displaystyle \frac{1}{17}}$;

в) ${\displaystyle \frac{2x^2+8x-5}{x+3}-2x=}$${\displaystyle 1}$;

${\displaystyle \frac{2x^2+8x-5}{x+3}-\frac{{2x}^{\mathrm{\setminus }x+3}}{1}-\frac{1^{\mathrm{\setminus }x+3}}{1}=}$${\displaystyle 0}$;

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

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

${\displaystyle \{\begin{array}{rl}& x-8=0\\ & x+3\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=8\\ & x\ne -3\end{array}}$;

${\displaystyle x=}$${\displaystyle 8}$;

г) ${\displaystyle \frac{3x}{x^2+4x}+\frac{x-4}{x^2+3x}=}$${\displaystyle \frac{9}{(x+3)(x+4)}}$;

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

${\displaystyle \frac{{3x}^{\mathrm{\setminus }x+3}}{x(x+4)}+\frac{{x-4}^{\mathrm{\setminus }x+4}}{x(x+3)}-\frac{9^{\mathrm{\setminus }x}}{(x+3)(x+4)}}$=

=${\displaystyle \frac{3x(x+3)+(x-4)(x+4)-9x}{x(x+3)(x+4)}}$=

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

${\displaystyle \{\begin{array}{rl}& 4(x-2)(x+2)=0\\ & x(x+3)(x+4)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-2\text{або }x=2\\ & x\ne -4,x\ne -3,x\ne 0\end{array}}$;

${\displaystyle x=}$${\displaystyle -2\text{або }x=}$${\displaystyle 2}$.