852 Вправа 852Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 255

852. Розв’яжіть рівняння:

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

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

${\displaystyle t^2+5t-6=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle -5}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle -6}$;

${\displaystyle t_1=}$${\displaystyle -6}$; ${\displaystyle t_2=}$${\displaystyle 1}$;

${\displaystyle x^2-x=}$${\displaystyle -6}$ або ${\displaystyle x^2-x=}$${\displaystyle 1}$;

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

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot 6=}$${\displaystyle 1-24=}$${\displaystyle -23<0}$ — коренів немає;

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

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-1)=}$${\displaystyle 1+4=}$${\displaystyle 5}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{1\pm \sqrt{5}}{2}}$;

2) ${\displaystyle 2{\left(\frac{x+6}{x}\right)}^2-11\left(\frac{x+6}{x}\right)+15=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle \frac{x+6}{x}}$, ${\displaystyle x\ne 0}$;

${\displaystyle 2t^2-11t+15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-11)}^2-4\cdot 2\cdot 15=}$${\displaystyle 121-120=}$${\displaystyle 1}$;

${\displaystyle t_1=}$${\displaystyle \frac{11-1}{4}=}$${\displaystyle 2,5}$; ${\displaystyle t_2=}$${\displaystyle \frac{11+1}{4}=}$${\displaystyle 3}$;

${\displaystyle \frac{x+6}{x}=}$${\displaystyle 2,5}$; ${\displaystyle x+6=}$${\displaystyle 2,5x}$; ${\displaystyle 1,5x=}$${\displaystyle 6}$; ${\displaystyle x=}$${\displaystyle 4}$;

${\displaystyle \frac{x+6}{x}=}$${\displaystyle 3}$; ${\displaystyle x+6=}$${\displaystyle 3x}$; ${\displaystyle 2x=}$${\displaystyle 6}$; ${\displaystyle x=}$${\displaystyle 3}$;

${\displaystyle x_1=}$${\displaystyle 4}$; ${\displaystyle x_2=}$${\displaystyle 3}$;

3) ${\displaystyle \frac{x+10}{x}-12\cdot \frac{x}{x+10}+1=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle \frac{x+10}{x}}$, ${\displaystyle x\ne 0}$, ${\displaystyle x\ne -10}$;

${\displaystyle t-\frac{12}{t}+1=}$${\displaystyle 0}$;

${\displaystyle \frac{t^2+t-12}{t}=}$${\displaystyle 0}$;

${\displaystyle t^2+t-12=}$${\displaystyle 0}$;

${\displaystyle t_1=}$${\displaystyle -4}$; ${\displaystyle t_2=}$${\displaystyle 3}$;

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

${\displaystyle \frac{x+10}{x}=}$${\displaystyle 3}$; ${\displaystyle x+10=}$${\displaystyle 3x}$; ${\displaystyle 2x=}$${\displaystyle 10}$; ${\displaystyle x=}$${\displaystyle 5}$;

${\displaystyle x_1=}$${\displaystyle -2}$; ${\displaystyle x_2=}$${\displaystyle 5}$;

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

${\displaystyle t=}$${\displaystyle \frac{x-3}{x+5}}$, ${\displaystyle x\ne -5}$, ${\displaystyle x\ne 3}$;

${\displaystyle t+\frac{1}{t}+2=}$${\displaystyle 0}$;

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

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

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

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

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

${\displaystyle 2x=}$${\displaystyle -2}$; ${\displaystyle x=}$${\displaystyle -1}$.