386 Вправа 386Алгебра 8 клас Бевз 2025 · § 8 · стор. 78

386. Розв’яжи рівняння.

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

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

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

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

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

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

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

б) ${\displaystyle \frac{6x^2+5}{3x+2}=}$${\displaystyle \frac{6x^2+x}{3x+2}}$;

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

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

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

${\displaystyle \{\begin{array}{rl}& x=5\\ & x\ne -\frac{2}{3}\end{array}}$;

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

в) ${\displaystyle \frac{4x^2-x}{2x+3}=}$${\displaystyle \frac{4x^2-9}{2x+3}}$;

${\displaystyle \frac{4x^2-x}{2x+3}-\frac{4x^2-9}{2x+3}=}$${\displaystyle 0}$;

${\displaystyle \frac{4x^2-x-(4x^2-9)}{2x+3}=}$${\displaystyle \frac{4x^2-x-4x^2+9}{2x+3}=}$${\displaystyle \frac{9-x}{2x+3}}$=0;

${\displaystyle \{\begin{array}{rl}& 9-x=0\\ & 2x+3\ne 0\end{array}}$;

${\displaystyle \{\begin{array}{rl}& x=9\\ & x\ne -\frac{3}{2}\end{array}}$;

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

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

допустимі ${\displaystyle x\ne 0,x\ne -2}$;

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

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

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

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

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

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

ґ) ${\displaystyle \frac{c+1}{c-1}=}$${\displaystyle \frac{c-5}{c-3}}$;

допустимі ${\displaystyle c\ne 1,c\ne 3}$;

${\displaystyle (c+1)(c-3)=}$${\displaystyle (c-5)(c-1)}$;

${\displaystyle c^2-3c+c-3=}$${\displaystyle c^2-c-5c+5}$;

${\displaystyle c^2-2c-3=}$${\displaystyle c^2-6c+5}$;

${\displaystyle -2c+6c=}$${\displaystyle 5+3}$;

${\displaystyle 4c=}$${\displaystyle 8}$;

${\displaystyle c=}$${\displaystyle \frac{8}{4}}$;

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

д) ${\displaystyle \frac{x^2-6x+9}{x^2-9}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& {(x-3)}^2=0,\\ & (x-3)(x+3)\ne 0;\end{array}}$ ${\displaystyle \{\begin{array}{rl}& x=3,\\ & x\ne 3,x\ne -3;\end{array}}$ — коренів немає.