370 Вправа 370Алгебра 8 клас Бевз 2025 · § 8 · стор. 76

370. Розв’яжи рівняння.

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

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

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

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

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

б) ${\displaystyle \frac{x}{x+6}+\frac{2^{\mathrm{\setminus }x+6}}{1}=}$${\displaystyle 0}$;

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

${\displaystyle \{\begin{array}{rl}& 3x+12=0\\ & x+6\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-4\\ & x\ne -6\end{array}}$;

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

в) ${\displaystyle \frac{3x-7}{x}+\frac{4^{\mathrm{\setminus }x}}{1}=}$${\displaystyle 0}$;

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

${\displaystyle \{\begin{array}{rl}& 7x-7=0\\ & x\ne 0\end{array}}$;

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

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

г) ${\displaystyle \frac{3x-4}{x}=}$${\displaystyle 2}$;

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

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

${\displaystyle \{\begin{array}{rl}& x-4=0\\ & x\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=4\\ & x\ne 0\end{array}}$;

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

ґ) ${\displaystyle \frac{5-2x}{x}=}$${\displaystyle 3}$;

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

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

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

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

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

д) ${\displaystyle \frac{2x}{x+3}=}$${\displaystyle 1}$;

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

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

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

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

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