921 Вправа 921Алгебра 8 клас Бевз 2025 · § 18 · стор. 175

921. Розв’яжи рівняння.

Зводимо дроби до спільного знаменника (стежимо, щоб знаменник не дорівнював нулю).

а) ${\displaystyle \frac{x^{\mathrm{\setminus }x-3}}{x+3}+\frac{x^{\mathrm{\setminus }x+3}}{x-3}=}$${\displaystyle 4,x\ne \pm 3}$;

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

${\displaystyle 2x^2=}$${\displaystyle 4(x^2-9)}$;

${\displaystyle 2x^2=}$${\displaystyle 4x^2-36}$;

${\displaystyle 2x^2=}$${\displaystyle 36}$;

${\displaystyle x^2=}$${\displaystyle 18}$;

${\displaystyle x=}$${\displaystyle \pm 3\sqrt{2}}$;

б) ${\displaystyle \frac{{x+2}^{\mathrm{\setminus }x+2}}{x-2}+\frac{{x-2}^{\mathrm{\setminus }x-2}}{x+2}=}$${\displaystyle \frac{10}{3},x\ne \pm 2}$;

${\displaystyle \frac{x^2+4x+4+x^2-4x+4}{x^2-4}=}$${\displaystyle \frac{10}{3}}$;

${\displaystyle 3(2x^2+8)=}$${\displaystyle 10(x^2-4)}$;

${\displaystyle 6x^2+24=}$${\displaystyle 10x^2-40}$;

${\displaystyle 4x^2=}$${\displaystyle 64}$;

${\displaystyle x^2=}$${\displaystyle 16}$;

${\displaystyle x=}$${\displaystyle \pm 4}$;

в) ${\displaystyle \frac{12}{x^2-4}+\frac{x^{\mathrm{\setminus }x+2}}{x-2}=}$${\displaystyle -3,x\ne \pm 2}$;

${\displaystyle 12+x(x+2)=}$${\displaystyle -3(x^2-4)}$;

${\displaystyle 12+x^2+2x=}$${\displaystyle -3x^2+12}$;

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

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

${\displaystyle x_1=}$${\displaystyle 0,x_2=}$${\displaystyle -0,5}$;

г) ${\displaystyle \frac{7(2x+5)}{{(x-5)}^2}+\frac{2^{\mathrm{\setminus }x-5}}{x-5}=}$${\displaystyle 1,x\ne 5}$;

${\displaystyle 14x+35+2x-10=}$${\displaystyle {(x-5)}^2}$;

${\displaystyle 16x+25=}$${\displaystyle x^2-10x+25}$;

${\displaystyle x^2-26x=}$${\displaystyle 0}$;

${\displaystyle x(x-26)=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle 0,x_2=}$${\displaystyle 26}$.