255 Вправа 255Алгебра 8 клас Тарасенкова 2025 · § 7 · стор. 76

255. Розв’яжіть рівняння:

1) ${\displaystyle 4x+\frac{3x-5x^2}{5x^2+9}+2=}$${\displaystyle 4x+1}$;

ОДЗ: ${\displaystyle 5x^2+9\ne 0}$ — виконується для будь-якого ${\displaystyle x}$, бо ${\displaystyle 5x^2+9>0}$;

${\displaystyle 4x+\frac{3x-5x^2}{5x^2+9}+2=}$${\displaystyle 4x+1}$;

${\displaystyle \frac{3x-5x^2}{5x^2+9}=}$${\displaystyle 1-2}$;

${\displaystyle \frac{3x-5x^2}{5x^2+9}=}$${\displaystyle -1}$;

${\displaystyle 3x-5x^2=}$${\displaystyle -1\cdot (5x^2+9)}$;

${\displaystyle 3x-5x^2=}$${\displaystyle -5x^2-9}$;

${\displaystyle 3x=}$${\displaystyle -9}$ | ${\displaystyle :3}$

${\displaystyle x=}$${\displaystyle -3}$;

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

${\displaystyle x^2-x-6=}$${\displaystyle 0}$: ${\displaystyle x_1=}$${\displaystyle 3}$, ${\displaystyle x_2=}$${\displaystyle -2}$; ${\displaystyle x^2-x-6=}$${\displaystyle (x-3)(x+2)}$;

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

ОДЗ: ${\displaystyle (x-3)(x+2)\ne 0}$;

${\displaystyle x\ne 3}$; ${\displaystyle x\ne -2}$;

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

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

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

${\displaystyle x+2-3x+9=}$${\displaystyle 2x}$;

${\displaystyle -2x+11=}$${\displaystyle 2x}$;

${\displaystyle -2x-2x=}$${\displaystyle -11}$;

${\displaystyle -4x=}$${\displaystyle -11}$ | ${\displaystyle :\left(-4\right)}$

${\displaystyle x=}$${\displaystyle \frac{11}{4}}$;

${\displaystyle x=}$${\displaystyle 2,75}$;

3) ${\displaystyle \frac{x}{3-x}+\frac{x+1}{x+3}=}$${\displaystyle \frac{15}{x^2-9}}$;

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

ОДЗ: ${\displaystyle (x-3)(x+3)\ne 0}$;

${\displaystyle x\ne 3}$; ${\displaystyle x\ne -3}$;

${\displaystyle -\frac{x^{\mathrm{\setminus }x+3}}{x-3}+\frac{{x+1}^{\mathrm{\setminus }x-3}}{x+3}=}$${\displaystyle \frac{15}{(x-3)(x+3)}}$;

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

${\displaystyle -x(x+3)+(x+1)(x-3)=}$${\displaystyle 15}$;

${\displaystyle -x^2-3x+x^2-3x+x-3=}$${\displaystyle 15}$;

${\displaystyle -5x=}$${\displaystyle 15+3}$;

${\displaystyle -5x=}$${\displaystyle 18}$ | ${\displaystyle :\left(-5\right)}$

${\displaystyle x=}$${\displaystyle -\frac{18}{5}}$;

${\displaystyle x=}$${\displaystyle -3,6}$;

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

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

ОДЗ: ${\displaystyle (x-2)(x+1)\ne 0}$;

${\displaystyle x\ne 2}$; ${\displaystyle x\ne -1}$;

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

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

${\displaystyle -4(x+1)+3(x-2)=}$${\displaystyle 3x}$;

${\displaystyle -4x-4+3x-6=}$${\displaystyle 3x}$;

${\displaystyle -x-10=}$${\displaystyle 3x}$;

${\displaystyle -x-3x=}$${\displaystyle 10}$;

${\displaystyle -4x=}$${\displaystyle 10}$ | ${\displaystyle :\left(-4\right)}$

${\displaystyle x=}$${\displaystyle -\frac{10}{4}=}$${\displaystyle -\frac{5}{2}}$;

${\displaystyle x=}$${\displaystyle -2,5}$.