844° Вправа 844Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 254

844. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x^2-4}{x-1}=}$${\displaystyle 4}$;

${\displaystyle \frac{x^2-4-4(x-1)}{x-1}=}$${\displaystyle 0}$;

${\displaystyle \frac{x^2-4-4x+4}{x-1}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& x^2-4x=0,\\ & x-1\ne 0\end{array}}$;

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

${\displaystyle x_1=}$${\displaystyle 0}$; ${\displaystyle x_2=}$${\displaystyle 4}$;

${\displaystyle x\ne 1}$;

${\displaystyle x_1=}$${\displaystyle 0}$; ${\displaystyle x_2=}$${\displaystyle 4}$;

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

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

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

${\displaystyle \{\begin{array}{rl}& x^2-3x-10=0,\\ & x-2\ne 0\end{array}}$;

${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{3\pm \sqrt{49}}{2\cdot 1}=}$${\displaystyle \frac{3\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+7}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-7}{2}=}$${\displaystyle -2}$;

${\displaystyle x\ne 2}$;

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

3) ${\displaystyle \frac{2}{x}+15=}$${\displaystyle \frac{8}{x^2}}$;

${\displaystyle \frac{2x+15x^2-8}{x^2}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& 15x^2+2x-8=0,\\ & x^2\ne 0\end{array}}$;

${\displaystyle D=}$${\displaystyle 2^2-4\cdot 15\cdot (-8)=}$${\displaystyle 4+480=}$${\displaystyle 484}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{484}}{2\cdot 15}=}$${\displaystyle \frac{-2\pm 22}{30}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+22}{30}=}$${\displaystyle \frac{20}{30}=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-22}{30}=}$${\displaystyle \frac{-24}{30}=}$${\displaystyle -0,8}$;

${\displaystyle x\ne 0}$;

${\displaystyle x_1=}$${\displaystyle \frac{2}{3}}$; ${\displaystyle x_2=}$${\displaystyle -0,8}$;

4) ${\displaystyle \frac{1}{x}+1=}$${\displaystyle \frac{20}{x^2}}$;

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

${\displaystyle \{\begin{array}{rl}& x^2+x-20=0,\\ & x^2\ne 0\end{array}}$;

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-20)=}$${\displaystyle 1+80=}$${\displaystyle 81}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-1\pm \sqrt{81}}{2\cdot 1}=}$${\displaystyle \frac{-1\pm 9}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+9}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-9}{2}=}$${\displaystyle -5}$;

${\displaystyle x\ne 0}$;

${\displaystyle x_1=}$${\displaystyle 4}$; ${\displaystyle x_2=}$${\displaystyle -5}$.

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

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

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

${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 2\cdot (-5)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{3\pm \sqrt{49}}{2\cdot 2}=}$${\displaystyle \frac{3\pm 7}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+7}{4}=}$${\displaystyle \frac{10}{4}=}$${\displaystyle 2,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{3-7}{4}=}$${\displaystyle -1}$;

${\displaystyle x\ne 0}$;

${\displaystyle x_1=}$${\displaystyle 2,5}$; ${\displaystyle x_2=}$${\displaystyle -1}$;

6) ${\displaystyle \frac{3x^2+x-24}{x^2-9}=}$${\displaystyle 2}$;

${\displaystyle \frac{3x^2+x-24-2(x^2-9)}{x^2-9}=}$${\displaystyle 0}$;

${\displaystyle \frac{3x^2+x-24-2x^2+18}{x^2-9}=}$${\displaystyle 0}$;

${\displaystyle \{\begin{array}{rl}& x^2+x-6=0,\\ & x^2-9\ne 0\end{array}}$;

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-6)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-1\pm \sqrt{25}}{2\cdot 1}=}$${\displaystyle \frac{-1\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+5}{2}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-5}{2}=}$${\displaystyle -3}$;

${\displaystyle x^2-9\ne 0}$;

${\displaystyle x^2\ne 9}$;

${\displaystyle x\ne 3}$, ${\displaystyle x\ne -3}$;

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