862* Вправа 862Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

862. Знайдіть найбільший корінь рівняння:

1) ${\displaystyle 5{(x+1)}^2-8|x+1|+3=}$${\displaystyle 0}$; оскільки ${\displaystyle {(x+1)}^2=}$${\displaystyle {|x+1|}^2}$, нехай ${\displaystyle |x+1|=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle 5t^2-8t+3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-8)}^2-4\cdot 5\cdot 3=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle t_1=}$${\displaystyle \frac{8-\sqrt{4}}{2\cdot 5}=}$${\displaystyle \frac{8-2}{10}=}$${\displaystyle 0,6}$;

${\displaystyle t_2=}$${\displaystyle \frac{8+\sqrt{4}}{2\cdot 5}=}$${\displaystyle \frac{8+2}{10}=}$${\displaystyle 1}$;

${\displaystyle |x+1|=}$${\displaystyle 0,6}$ або ${\displaystyle |x+1|=}$${\displaystyle 1}$;

${\displaystyle x+1=}$${\displaystyle \pm 0,6}$ або ${\displaystyle x+1=}$${\displaystyle \pm 1}$;

${\displaystyle x_1=}$${\displaystyle -0,4}$; ${\displaystyle x_2=}$${\displaystyle -1,6}$; ${\displaystyle x_3=}$${\displaystyle 0}$; ${\displaystyle x_4=}$${\displaystyle -2}$;

Найбільший корінь: ${\displaystyle 0}$;

2) ${\displaystyle \frac{4}{x^2+10\left|x\right|}-\frac{2}{5x}=}$${\displaystyle \frac{1}{25}}$; ОДЗ: ${\displaystyle x\ne 0}$;

якщо ${\displaystyle x>0}$, то ${\displaystyle \left|x\right|=}$${\displaystyle x}$:

${\displaystyle \frac{4^{\mathrm{\setminus }25}}{x(x+10)}-\frac{2^{\mathrm{\setminus }5\left(x+10\right)}}{5x}-\frac{1^{\mathrm{\setminus }x\left(x+10\right)}}{25}=}$${\displaystyle 0}$;

${\displaystyle \frac{4\cdot 25-2\cdot 5(x+10)-1\cdot x(x+10)}{25x(x+10)}=}$${\displaystyle 0}$;

${\displaystyle 100-10x-100-x^2-10x=}$${\displaystyle 0}$;

${\displaystyle -x^2-20x=}$${\displaystyle 0}$;

${\displaystyle -x(x+20)=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0}$ (не задовольняє ОДЗ) або ${\displaystyle x=}$${\displaystyle -20}$ (не задовольняє умову ${\displaystyle x>0}$);

якщо ${\displaystyle x<0}$, то ${\displaystyle \left|x\right|=}$${\displaystyle -x}$:

${\displaystyle \frac{4^{\mathrm{\setminus }25}}{x(x-10)}-\frac{2^{\mathrm{\setminus }5\left(x-10\right)}}{5x}-\frac{1^{\mathrm{\setminus }x\left(x-10\right)}}{25}=}$${\displaystyle 0}$;

${\displaystyle \frac{4\cdot 25-2\cdot 5(x-10)-1\cdot x(x-10)}{25x(x-10)}=}$${\displaystyle 0}$;

${\displaystyle 100-10x+100-x^2+10x=}$${\displaystyle 0}$;

${\displaystyle -x^2+200=}$${\displaystyle 0}$;

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

${\displaystyle x_1=}$${\displaystyle \sqrt{200}=}$${\displaystyle 10\sqrt{2}}$ (не задовольняє умову ${\displaystyle x<0}$);

${\displaystyle x_2=}$${\displaystyle -10\sqrt{2}}$;

Найбільший корінь: ${\displaystyle -10\sqrt{2}}$.