836° Вправа 836Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 253

836. Розв’яжіть рівняння:

1) ${\displaystyle x(x-1)=}$${\displaystyle 12}$;

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

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-12)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

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

${\displaystyle x_1=}$${\displaystyle \frac{1+7}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-7}{2}=}$${\displaystyle -3}$;

2) ${\displaystyle 9x(x+1)=}$${\displaystyle -x-1}$;

${\displaystyle 9x^2+9x+x+1=}$${\displaystyle 0}$;

${\displaystyle 9x^2+10x+1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {10}^2-4\cdot 9\cdot 1=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-10\pm \sqrt{64}}{2\cdot 9}=}$${\displaystyle \frac{-10\pm 8}{18}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-10+8}{18}=}$${\displaystyle \frac{-2}{18}=}$${\displaystyle -\frac{1}{9}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-10-8}{18}=}$${\displaystyle -1}$;

3) ${\displaystyle {(x+4)}^2+x^2=}$${\displaystyle x+31}$;

${\displaystyle x^2+8x+16+x^2-x-31=}$${\displaystyle 0}$;

${\displaystyle 2x^2+7x-15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 7^2-4\cdot 2\cdot (-15)=}$${\displaystyle 49+120=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-7\pm \sqrt{169}}{2\cdot 2}=}$${\displaystyle \frac{-7\pm 13}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-7+13}{4}=}$${\displaystyle \frac{6}{4}=}$${\displaystyle 1,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-7-13}{4}=}$${\displaystyle -5}$.

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

${\displaystyle x^2-2x+1+x^2+2x+1-5x=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 2=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{9}}{2\cdot 2}=}$${\displaystyle \frac{5\pm 3}{4}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{5-3}{4}=}$${\displaystyle \frac{2}{4}=}$${\displaystyle 0,5}$;

5) ${\displaystyle {(x+3)}^2+{(x-3)}^2=}$${\displaystyle 9(x+1)}$;

${\displaystyle x^2+6x+9+x^2-6x+9-9x-9=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-9)}^2-4\cdot 2\cdot 9=}$${\displaystyle 81-72=}$${\displaystyle 9}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{9\pm \sqrt{9}}{2\cdot 2}=}$${\displaystyle \frac{9\pm 3}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{9+3}{4}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{9-3}{4}=}$${\displaystyle \frac{6}{4}=}$${\displaystyle 1,5}$;

6) ${\displaystyle {(3x-2)}^2+11=}$${\displaystyle 4x(2x-1)}$;

${\displaystyle 9x^2-12x+4+11=}$${\displaystyle 8x^2-4x}$;

${\displaystyle 9x^2-8x^2-12x+4x+15=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot 15=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{8\pm \sqrt{4}}{2\cdot 1}=}$${\displaystyle \frac{8\pm 2}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{8+2}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-2}{2}=}$${\displaystyle 3}$.