387 Вправа 387Алгебра 8 клас Бевз 2025 · § 8 · стор. 78

387. Розв’яжи рівняння.

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

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

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

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

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

${\displaystyle x=}$${\displaystyle 2}$;

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

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

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

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

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

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

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

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

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

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

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

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

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

допустимі ${\displaystyle x\ne 0,x\ne -3}$;

${\displaystyle 2\cdot 4x^2=}$${\displaystyle (3-x)(3+x)}$;

${\displaystyle 8x^2=}$${\displaystyle 9-x^2}$;

${\displaystyle 8x^2+x^2=}$${\displaystyle 9}$;

${\displaystyle 9x^2=}$${\displaystyle 9}$;

${\displaystyle x^2=}$${\displaystyle 1}$;

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

ґ) ${\displaystyle \frac{y-2}{y-6}=}$${\displaystyle \frac{y}{y-5}}$;

допустимі ${\displaystyle y\ne 5,y\ne 6}$;

${\displaystyle (y-2)(y-5)=}$${\displaystyle y(y-6)}$;

${\displaystyle y^2-5y-2y+10=}$${\displaystyle y^2-6y}$;

${\displaystyle y^2-7y+10=}$${\displaystyle y^2-6y}$;

${\displaystyle -7y+6y=}$${\displaystyle -10}$;

${\displaystyle -y=}$${\displaystyle -10}$;

${\displaystyle y=}$${\displaystyle 10}$;

д) ${\displaystyle \frac{z^2-4z+4}{z^2-4}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& {(z-2)}^2=0,\\ & (z-2)(z+2)\ne 0;\end{array}}$ ${\displaystyle \{\begin{array}{rl}& z=2,\\ & z\ne 2,z\ne -2;\end{array}}$ — коренів немає.