372 Вправа 372Алгебра 8 клас Бевз 2025 · § 8 · стор. 77

372. Розв’яжи рівняння.

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

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

${\displaystyle \frac{3x+5(x+2)}{x+2}=}$${\displaystyle \frac{3x+5x+10}{x+2}=}$${\displaystyle \frac{8x+10}{x+2}}$=0;

${\displaystyle \{\begin{array}{rl}& 8x+10=0\\ & x+2\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{5}{4}\\ & x\ne -2\end{array}}$;

${\displaystyle x=}$${\displaystyle -1,25}$;

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

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

${\displaystyle \frac{2x+5-3(x+2)}{x+2}=}$${\displaystyle \frac{2x+5-3x-6}{x+2}=}$${\displaystyle \frac{-x-1}{x+2}}$=0;

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

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

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

в) ${\displaystyle \frac{x-6}{3x-1}=}$${\displaystyle 5}$;

${\displaystyle \frac{x-6}{3x-1}-\frac{5^{\mathrm{\setminus }3x-1}}{1}=}$${\displaystyle 0}$;

${\displaystyle \frac{x-6-5(3x-1)}{3x-1}=}$${\displaystyle \frac{x-6-15x+5}{3x-1}=}$${\displaystyle \frac{-14x-1}{3x-1}}$=0;

${\displaystyle \{\begin{array}{rl}& -14x-1=0\\ & 3x-1\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{1}{14}\\ & x\ne \frac{1}{3}\end{array}}$;

${\displaystyle x=}$${\displaystyle -\frac{1}{14}}$;

г) ${\displaystyle \frac{7x^2+1}{2x}=}$${\displaystyle 4x}$;

${\displaystyle \frac{7x^2+1}{2x}-\frac{{4x}^{\mathrm{\setminus }2x}}{1}=}$${\displaystyle 0}$;

${\displaystyle \frac{7x^2+1-8x^2}{2x}=}$${\displaystyle \frac{1-x^2}{2x}}$=0;

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

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

${\displaystyle x=}$${\displaystyle -1\text{або }x=}$${\displaystyle 1}$;

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

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

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

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

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

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

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

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

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

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

${\displaystyle x=}$${\displaystyle -1\text{або }x=}$${\displaystyle 1}$;

${\displaystyle x=}$${\displaystyle -1\text{або }x=}$${\displaystyle 1}$.