768 Вправа 768Алгебра 8 клас Тарасенкова 2025 · § 18 · стор. 227

768. За яких значень змінної х:

1) ${\displaystyle (-x-1)(x-4)=}$${\displaystyle x(4x-11)}$;

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

${\displaystyle -x^2+3x+4-4x^2+11x=}$${\displaystyle 0}$;

${\displaystyle -5x^2+14x+4=}$${\displaystyle 0}$ | ${\displaystyle \cdot \left(-1\right)}$

${\displaystyle 5x^2-14x-4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-14)}^2-4\cdot 5\cdot (-4)=}$${\displaystyle 196+80=}$${\displaystyle 276}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{14\pm \sqrt{276}}{2\cdot 5}=}$${\displaystyle \frac{14\pm 2\sqrt{69}}{10}=}$${\displaystyle \frac{7\pm \sqrt{69}}{5}}$;

${\displaystyle x_1=}$${\displaystyle \frac{7+\sqrt{69}}{5}}$;

${\displaystyle x_2=}$${\displaystyle \frac{7-\sqrt{69}}{5}}$;

2) ${\displaystyle x^2-4,6x+2,4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-4,6)}^2-4\cdot 1\cdot 2,4=}$${\displaystyle 21,16-9,6=}$${\displaystyle 11,56}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{4,6\pm \sqrt{11,56}}{2\cdot 1}=}$${\displaystyle \frac{4,6\pm 3,4}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{4,6+3,4}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{4,6-3,4}{2}=}$${\displaystyle 0,6}$;

3) ${\displaystyle x^2-2x=}$${\displaystyle 28-5x}$;

${\displaystyle x^2-2x-28+5x=}$${\displaystyle 0}$;

${\displaystyle x^2+3x-28=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 3^2-4\cdot 1\cdot (-28)=}$${\displaystyle 9+112=}$${\displaystyle 121}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-3\pm \sqrt{121}}{2\cdot 1}=}$${\displaystyle \frac{-3\pm 11}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+11}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-11}{2}=}$${\displaystyle -7}$.