762 Вправа 762Алгебра 8 клас Тарасенкова 2025 · § 18 · стор. 227

762. Розв’яжіть рівняння:

1) ${\displaystyle x(x+1)=}$${\displaystyle 56}$;

${\displaystyle x^2+x-56=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-56)=}$${\displaystyle 1+224=}$${\displaystyle 225}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-1\pm \sqrt{225}}{2\cdot 1}=}$${\displaystyle \frac{-1\pm 15}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+15}{2}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-15}{2}=}$${\displaystyle -8}$;

2) ${\displaystyle (y-2)(2y-1)=}$${\displaystyle 5}$;

${\displaystyle 2y^2-y-4y+2-5=}$${\displaystyle 0}$;

${\displaystyle 2y^2-5y-3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot (-3)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{49}}{2\cdot 2}=}$${\displaystyle \frac{5\pm 7}{4}}$;

${\displaystyle y_1=}$${\displaystyle \frac{5+7}{4}=}$${\displaystyle 3}$;

${\displaystyle y_2=}$${\displaystyle \frac{5-7}{4}=}$${\displaystyle \frac{-2}{4}=}$${\displaystyle -0,5}$;

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

${\displaystyle 12-8z=}$${\displaystyle z^2-6z+9}$;

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

${\displaystyle D=}$${\displaystyle 2^2-4\cdot 1\cdot (-3)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{16}}{2\cdot 1}=}$${\displaystyle \frac{-2\pm 4}{2}}$;

${\displaystyle z_1=}$${\displaystyle \frac{-2+4}{2}=}$${\displaystyle 1}$;

${\displaystyle z_2=}$${\displaystyle \frac{-2-4}{2}=}$${\displaystyle -3}$;

4) ${\displaystyle (x+6)(x-2)=}$${\displaystyle 2(x-2)}$;

${\displaystyle x^2-2x+6x-12=}$${\displaystyle 2x-4}$;

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

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

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{36}}{2\cdot 1}=}$${\displaystyle \frac{-2\pm 6}{2}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{-2-6}{2}=}$${\displaystyle -4}$.