761 Вправа 761Алгебра 8 клас Тарасенкова 2025 · § 18 · стор. 226

761. Розв’яжіть рівняння:

1) ${\displaystyle x(x-1)=}$${\displaystyle 72}$;

${\displaystyle x^2-x-72=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-72)=}$${\displaystyle 1+288=}$${\displaystyle 289}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{1\pm \sqrt{289}}{2\cdot 1}=}$${\displaystyle \frac{1\pm 17}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+17}{2}=}$${\displaystyle 9}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-17}{2}=}$${\displaystyle -8}$;

2) ${\displaystyle 2y(y+2)-3=}$${\displaystyle 9y}$;

${\displaystyle 2y^2+4y-3-9y=}$${\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 (2z-3)(2-3z)=}$${\displaystyle -4}$;

${\displaystyle 4z-6z^2-6+9z+4=}$${\displaystyle 0}$;

${\displaystyle -6z^2+13z-2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {13}^2-4\cdot (-6)\cdot (-2)=}$${\displaystyle 169-48=}$${\displaystyle 121}$;

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-13\pm \sqrt{121}}{2\cdot (-6)}=}$${\displaystyle \frac{-13\pm 11}{-12}}$;

${\displaystyle z_1=}$${\displaystyle \frac{-13+11}{-12}=}$${\displaystyle \frac{-2}{-12}=}$${\displaystyle \frac{1}{6}}$;

${\displaystyle z_2=}$${\displaystyle \frac{-13-11}{-12}=}$${\displaystyle 2}$;

4) ${\displaystyle {(x+5)}^2=}$${\displaystyle 4(x+10)}$;

${\displaystyle x^2+10x+25=}$${\displaystyle 4x+40}$;

${\displaystyle x^2+6x-15=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 6^2-4\cdot 1\cdot (-15)=}$${\displaystyle 36+60=}$${\displaystyle 96}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-6\pm \sqrt{96}}{2\cdot 1}=}$${\displaystyle \frac{-6\pm 4\sqrt{6}}{2}=}$${\displaystyle -3\pm 2\sqrt{6}}$;

${\displaystyle x_1=}$${\displaystyle -3+2\sqrt{6}}$;

${\displaystyle x_2=}$${\displaystyle -3-2\sqrt{6}}$;

5) ${\displaystyle 2y(y+3)=}$${\displaystyle {(3+y)}^2}$;

${\displaystyle 2y^2+6y=}$${\displaystyle 9+6y+y^2}$;

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

${\displaystyle y^2=}$${\displaystyle 9}$;

${\displaystyle y=}$${\displaystyle \sqrt{9}}$ або ${\displaystyle y=}$${\displaystyle -\sqrt{9}}$;

${\displaystyle y_1=}$${\displaystyle 3}$ або ${\displaystyle y_2=}$${\displaystyle -3}$;

6) ${\displaystyle {(z+4)}^2+{(z-4)}^2=}$${\displaystyle 36}$;

${\displaystyle z^2+8z+16+z^2-8z+16=}$${\displaystyle 36}$;

${\displaystyle 2z^2+32=}$${\displaystyle 36}$;

${\displaystyle 2z^2=}$${\displaystyle 4}$ | ${\displaystyle :2}$

${\displaystyle z^2=}$${\displaystyle 2}$;

${\displaystyle z_1=}$${\displaystyle \sqrt{2}}$ або ${\displaystyle z_2=}$${\displaystyle -\sqrt{2}}$;

7) ${\displaystyle (x+2)(x+1)=}$${\displaystyle 4(x^2-22)}$;

${\displaystyle x^2+3x+2=}$${\displaystyle 4x^2-88}$;

${\displaystyle 4x^2-88-x^2-3x-2=}$${\displaystyle 0}$;

${\displaystyle 3x^2-3x-90=}$${\displaystyle 0}$ | ${\displaystyle :3}$

${\displaystyle x^2-x-30=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-30)=}$${\displaystyle 1+120=}$${\displaystyle 121}$;

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

${\displaystyle x_1=}$${\displaystyle \frac{1+11}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-11}{2}=}$${\displaystyle -5}$;

8) ${\displaystyle {(y+2)}^2-10=}$${\displaystyle 6(y+3)}$;

${\displaystyle y^2+4y+4-10=}$${\displaystyle 6y+18}$;

${\displaystyle y^2-2y-24=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-24)=}$${\displaystyle 4+96=}$${\displaystyle 100}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{2\pm \sqrt{100}}{2\cdot 1}=}$${\displaystyle \frac{2\pm 10}{2}}$;

${\displaystyle y_1=}$${\displaystyle \frac{2+10}{2}=}$${\displaystyle 6}$;

${\displaystyle y_2=}$${\displaystyle \frac{2-10}{2}=}$${\displaystyle -4}$.