173 Вправа 173Алгебра 8 клас Бевз 2025 · § 4 · стор. 38

173. Розв’яжи рівняння.

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

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

=${\displaystyle \frac{15x-10-10x-5}{(2x+1)(3x-2)}=}$${\displaystyle \frac{5x-15}{(2x+1)(3x-2)}}$=0;

${\displaystyle \{\begin{array}{rl}& 5x-15=0\\ & (2x+1)(3x-2)\ne 0\end{array}}$;

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

${\displaystyle x=}$${\displaystyle 3}$;

б) ${\displaystyle \frac{3}{1+2x}+\frac{2}{1-x}=}$${\displaystyle 0}$;

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

=${\displaystyle \frac{3-3x+4x+2}{(1-x)(2x+1)}=}$${\displaystyle \frac{x+5}{(1-x)(2x+1)}}$=0;

${\displaystyle \{\begin{array}{rl}& x+5=0\\ & (1-x)(2x+1)\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-5\\ & x\ne -\frac{1}{2},x\ne 1\end{array}}$;

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

в) ${\displaystyle \frac{3x-7}{x-5}+\frac{x+3}{5-x}=}$${\displaystyle 0}$;

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

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

${\displaystyle \{\begin{array}{rl}& 2x-10=0\\ & x-5\ne 0\end{array}}$;

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

коренів немає;

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

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

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

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

${\displaystyle x=}$${\displaystyle 0,5}$.