203 Вправа 203Алгебра 8 клас Бевз 2025 · § 4 · стор. 41

203. Розв’яжи рівняння.

а) ${\displaystyle \frac{x-25}{5x-25}-\frac{3x+5}{5x-x^2}=}$${\displaystyle 0}$;

${\displaystyle \frac{{x-25}^{\mathrm{\setminus }x}}{5(x-5)}+\frac{{3x+5}^{\mathrm{\setminus }5}}{x(x-5)}=}$${\displaystyle \frac{x(x-25)+5(3x+5)}{5x(x-5)}}$=

=${\displaystyle \frac{x^2-25x+15x+25}{5x(x-5)}=}$${\displaystyle \frac{x^2-10x+25}{5x(x-5)}}$=0;

${\displaystyle \{\begin{array}{rl}& {(x-5)}^2=0\\ & 5x(x-5)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=5\\ & x\ne 0,x\ne 5\end{array}}$;

коренів немає;

б) ${\displaystyle \frac{x^2-6x}{x^2-4}+\frac{x}{x-2}=}$${\displaystyle 0}$;

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

=${\displaystyle \frac{x^2-6x+x^2+2x}{(x-2)(x+2)}=}$${\displaystyle \frac{2x^2-4x}{(x-2)(x+2)}}$=0;

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

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

${\displaystyle x=}$${\displaystyle 0}$;

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

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

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

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

чисельник ${\displaystyle x^2+4\ne 0}$ — коренів немає;

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

${\displaystyle \frac{1^{\mathrm{\setminus }4x^2+2x+1}}{2x-1}-\frac{6x}{(2x-1)(4x^2+2x+1)}}$=

=${\displaystyle \frac{4x^2+2x+1-6x}{(2x-1)(4x^2+2x+1)}=}$${\displaystyle \frac{4x^2-4x+1}{(2x-1)(4x^2+2x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& {(2x-1)}^2=0\\ & (2x-1)(4x^2+2x+1)\ne 0\end{array}}$;

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

коренів немає.