371 Вправа 371Алгебра 8 клас Бевз 2025 · § 8 · стор. 76

371. Розв’яжи рівняння.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

чисельник ${\displaystyle -x^2-1\ne 0}$ — коренів немає;

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

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

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

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

${\displaystyle x=}$${\displaystyle 6}$;

ґ) ${\displaystyle \frac{4x-2}{x+3}+\frac{6^{\mathrm{\setminus }x+3}}{1}=}$${\displaystyle 0}$;

${\displaystyle \frac{4x-2+6(x+3)}{x+3}=}$${\displaystyle \frac{4x-2+6x+18}{x+3}=}$${\displaystyle \frac{10x+16}{x+3}}$=0;

${\displaystyle \{\begin{array}{rl}& 10x+16=0\\ & x+3\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=-\frac{8}{5}\\ & x\ne -3\end{array}}$;

${\displaystyle x=}$${\displaystyle -1,6}$;

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

${\displaystyle \frac{4-3x-5(1-2x)}{1-2x}=}$${\displaystyle \frac{4-3x-5+10x}{1-2x}=}$${\displaystyle \frac{7x-1}{1-2x}}$=0;

${\displaystyle \{\begin{array}{rl}& 7x-1=0\\ & 1-2x\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=\frac{1}{7}\\ & x\ne \frac{1}{2}\end{array}}$;

${\displaystyle x=}$${\displaystyle \frac{1}{7}}$.