389 Вправа 389Алгебра 8 клас Бевз 2025 · § 8 · стор. 79

389. Знайди корені рівняння.

а) ${\displaystyle \frac{7x-2}{4-x^2}+\frac{5}{2+x}=}$${\displaystyle \frac{3}{2-x}}$;

${\displaystyle \frac{7x-2}{4-x^2}+\frac{5}{2+x}-\frac{3}{2-x}=}$${\displaystyle 0}$;

${\displaystyle \frac{7x-2}{(2-x)(x+2)}+\frac{5^{\mathrm{\setminus }2-x}}{x+2}-\frac{3^{\mathrm{\setminus }x+2}}{2-x}}$=

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

=${\displaystyle \frac{7x-2+10-5x-3x-6}{(2-x)(x+2)}=}$${\displaystyle \frac{2-x}{(2-x)(x+2)}}$=0;

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

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

коренів немає;

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

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

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

=${\displaystyle \frac{3x^2-14x+15-x^2-6x-5-2x^2+4x+6}{(x-3)(x+1)}}$=

=${\displaystyle \frac{16-16x}{(x-3)(x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& 16-16x=0\\ & (x-3)(x+1)\ne 0\end{array}}$;

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

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

в) ${\displaystyle 2x-\frac{4x^2-1}{2x-1}=}$${\displaystyle 5}$;

${\displaystyle \frac{{2x}^{\mathrm{\setminus }2x-1}}{1}-\frac{4x^2-1}{2x-1}-\frac{5^{\mathrm{\setminus }2x-1}}{1}=}$${\displaystyle 0}$;

${\displaystyle \frac{2x(2x-1)-(4x^2-1)-5(2x-1)}{2x-1}}$=

=${\displaystyle \frac{4x^2-2x-4x^2+1-10x+5}{2x-1}=}$${\displaystyle \frac{6-12x}{2x-1}}$=0;

${\displaystyle \{\begin{array}{rl}& 6-12x=0\\ & 2x-1\ne 0\end{array}}$;

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

коренів немає;

г) ${\displaystyle \frac{2x+2}{x^2-x}-\frac{x}{x^2-2x}=}$${\displaystyle -\frac{1}{(x-1)(x-2)}}$;

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

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

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

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

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

${\displaystyle \{\begin{array}{rl}& x=-2\text{або }x=2\\ & x\ne 0,x\ne 1,x\ne 2\end{array}}$;

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