46 Вправа 46 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

46. Розв’яжіть рівняння:

1) ${\displaystyle 5x(x-4)+1=}$${\displaystyle x^2-16x}$;

${\displaystyle 5x^2-20x+1-x^2+16x=}$${\displaystyle 0}$;

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

${\displaystyle D=}$${\displaystyle {(-4)}^2-4\cdot 4\cdot 1=}$${\displaystyle 16-16=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{4}{8}=}$${\displaystyle 0,5}$;

2) ${\displaystyle (y-1)(y+3)=}$${\displaystyle y+3}$;

${\displaystyle y^2+3y-y-3-y-3=}$${\displaystyle 0}$;

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

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

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

${\displaystyle y_1=}$${\displaystyle \frac{-1+5}{2}=}$${\displaystyle 2}$;

${\displaystyle y_2=}$${\displaystyle \frac{-1-5}{2}=}$${\displaystyle -3}$;

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

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

${\displaystyle x^2-6x+8=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot 8=}$${\displaystyle 36-32=}$${\displaystyle 4}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{6\pm \sqrt{4}}{2\cdot 1}=}$${\displaystyle \frac{6\pm 2}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{6+2}{2}=}$${\displaystyle 4}$;

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

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

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

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

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-4)=}$${\displaystyle 1+16=}$${\displaystyle 17}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-1\pm \sqrt{17}}{2\cdot 1}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+\sqrt{17}}{2}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-\sqrt{17}}{2}}$.