922 Вправа 922Алгебра 8 клас Бевз 2025 · § 18 · стор. 175

922. Розв’яжи рівняння.

а) ${\displaystyle \frac{{x-1}^{\mathrm{\setminus }x-1}}{x-2}+\frac{{x-2}^{\mathrm{\setminus }x-2}}{x-1}=}$${\displaystyle \frac{5}{2},x\ne 1,x\ne 2}$;

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

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

${\displaystyle 4x^2-12x+10=}$${\displaystyle 5x^2-15x+10}$;

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

${\displaystyle x(x-3)=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle 0,x_2=}$${\displaystyle 3}$;

б) ${\displaystyle \frac{x^{\mathrm{\setminus }x-5}}{x+5}+\frac{x^{\mathrm{\setminus }x+5}}{x-5}=}$${\displaystyle \frac{8}{3},x\ne \pm 5}$;

${\displaystyle \frac{2x^2}{x^2-25}=}$${\displaystyle \frac{8}{3}}$;

${\displaystyle 6x^2=}$${\displaystyle 8x^2-200}$;

${\displaystyle 2x^2=}$${\displaystyle 200}$;

${\displaystyle x^2=}$${\displaystyle 100}$;

${\displaystyle x=}$${\displaystyle \pm 10}$;

в) ${\displaystyle \frac{4}{3+x}=}$${\displaystyle \frac{3-4x^2}{x^2-9}-\frac{5}{x-3},x\ne \pm 3}$;

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

${\displaystyle 4x-12=}$${\displaystyle 3-4x^2-5x-15}$;

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

${\displaystyle x(4x+9)=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle 0,x_2=}$${\displaystyle -2,25}$;

г) ${\displaystyle \frac{5(x+2)}{{(x+3)}^2}+\frac{1^{\mathrm{\setminus }x+3}}{x+3}=}$${\displaystyle 1,x\ne -3}$;

${\displaystyle 5x+10+x+3=}$${\displaystyle {(x+3)}^2}$;

${\displaystyle 6x+13=}$${\displaystyle x^2+6x+9}$;

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

${\displaystyle x=}$${\displaystyle \pm 2}$.