854 Вправа 854Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 255

854. Розв’яжіть рівняння:

1) ${\displaystyle \frac{{10}^{\mathrm{\setminus }x}}{x-3}-1^{\mathrm{\setminus }x\left(x-3\right)}=}$${\displaystyle \frac{8^{\mathrm{\setminus }x-3}}{x}}$;

${\displaystyle \frac{10x-x(x-3)-8(x-3)}{x(x-3)}=}$${\displaystyle 0}$;

${\displaystyle \frac{10x-x^2+3x-8x+24}{x(x-3)}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& -x^2+5x+24=0,\\ & x\ne 0,x\ne 3\end{array}}$;

${\displaystyle x^2-5x-24=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle 5}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -24}$;

${\displaystyle x_1=}$${\displaystyle 8}$; ${\displaystyle x_2=}$${\displaystyle -3}$;

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

${\displaystyle \frac{x(x+3)-6(x+2)-x}{x(x+2)}=}$${\displaystyle 0}$;

${\displaystyle \frac{x^2+3x-6x-12-x}{x(x+2)}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& x^2-4x-12=0,\\ & x\ne 0,x\ne -2\end{array}}$;

${\displaystyle x_1+x_2=}$${\displaystyle 4}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -12}$;

${\displaystyle x_1=}$${\displaystyle 6}$; ${\displaystyle x_2=}$${\displaystyle -2}$;

${\displaystyle x_2=}$${\displaystyle -2}$ — не задовольняє умову ${\displaystyle x\ne -2}$;

${\displaystyle x=}$${\displaystyle 6}$;

3) ${\displaystyle \frac{1^{\mathrm{\setminus }8\left(x-3\right)}}{x+3}+\frac{1^{\mathrm{\setminus }8\left(x+3\right)}}{x-3}=}$${\displaystyle \frac{5^{\mathrm{\setminus }\left(x+3\right)\left(x-3\right)}}{8}}$;

${\displaystyle \frac{8(x-3)+8(x+3)-5(x^2-9)}{8(x^2-9)}=}$${\displaystyle 0}$;

${\displaystyle \frac{8x-24+8x+24-5x^2+45}{8(x^2-9)}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& -5x^2+16x+45=0,\\ & x\ne \pm 3\end{array}}$;

${\displaystyle 5x^2-16x-45=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-16)}^2-4\cdot 5\cdot (-45)=}$${\displaystyle 256+900=}$${\displaystyle 1156}$;

${\displaystyle x_1=}$${\displaystyle \frac{16+34}{10}=}$${\displaystyle 5}$; ${\displaystyle x_2=}$${\displaystyle \frac{16-34}{10}=}$${\displaystyle -1,8}$;

4) ${\displaystyle \frac{{x+3}^{\mathrm{\setminus }3\left(x+3\right)}}{x-3}+\frac{{x-3}^{\mathrm{\setminus }3\left(x-3\right)}}{x+3}=}$${\displaystyle \frac{{10}^{\mathrm{\setminus }\left(x-3\right)\left(x+3\right)}}{3}}$;

${\displaystyle \frac{3{(x+3)}^2+3{(x-3)}^2-10(x^2-9)}{3(x^2-9)}=}$${\displaystyle 0}$;

${\displaystyle \frac{3x^2+18x+27+3x^2-18x+27-10x^2+90}{3(x^2-9)}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& -4x^2+144=0,\\ & x\ne \pm 3\end{array}}$;

${\displaystyle 4x^2=}$${\displaystyle 144}$ | ${\displaystyle :4}$

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

${\displaystyle x_{1,2}=}$${\displaystyle \pm 6}$.