818 Вправа 818Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 243

818. Скоротіть дріб:

1)

${\displaystyle 3x^2+2x-1=}$${\displaystyle 0}$;

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

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{16}}{2\cdot 3}=}$${\displaystyle \frac{-2\pm 4}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+4}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-4}{6}=}$${\displaystyle -1}$;

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

${\displaystyle \frac{3x-1}{3x^2+2x-1}=}$${\displaystyle \frac{\cancel{\left(3x-1\right)}}{\cancel{\left(3x-1\right)}(x+1)}=}$${\displaystyle \frac{1}{x+1}}$;

2)

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

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

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

${\displaystyle x_1=}$${\displaystyle \frac{5+7}{4}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-7}{4}=}$${\displaystyle \frac{-2}{4}=}$${\displaystyle -0,5}$;

${\displaystyle 2x^2-5x-3=}$${\displaystyle 2(x-3)(x+0,5)=}$${\displaystyle (x-3)(2x+1)}$;

${\displaystyle \frac{2x^2-5x-3}{6x-18}=}$${\displaystyle \frac{\cancel{\left(x-3\right)}(2x+1)}{6\cancel{\left(x-3\right)}}=}$${\displaystyle \frac{2x+1}{6}}$;

3)

${\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 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-7}{2}=}$${\displaystyle -4}$;

${\displaystyle x^2+x-12=}$${\displaystyle (x+4)(x-3)}$;

${\displaystyle \frac{16-x^2}{x^2+x-12}=}$${\displaystyle \frac{(4-x)\cancel{\left(4+x\right)}}{(x-3)\cancel{\left(x+4\right)}}=}$${\displaystyle \frac{4-x}{x-3}=}$${\displaystyle \frac{x-4}{3-x}}$;

4)

${\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 8}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-15}{2}=}$${\displaystyle -7}$;

${\displaystyle x^2-x-56=}$${\displaystyle (x+7)(x-8)}$;

${\displaystyle \frac{x^2-x-56}{x^2+7x}=}$${\displaystyle \frac{\cancel{\left(x+7\right)}(x-8)}{x\cancel{\left(x+7\right)}}=}$${\displaystyle \frac{x-8}{x}}$;

5)

${\displaystyle x^2-3x-10=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{3\pm \sqrt{49}}{2\cdot 1}=}$${\displaystyle \frac{3\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+7}{2}=}$${\displaystyle 5}$;

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

${\displaystyle x^2-3x-10=}$${\displaystyle (x+2)(x-5)}$;

${\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}$;

${\displaystyle x^2-8x+15=}$${\displaystyle (x-3)(x-5)}$;

${\displaystyle \frac{x^2-3x-10}{x^2-8x+15}=}$${\displaystyle \frac{(x+2)\cancel{\left(x-5\right)}}{(x-3)\cancel{\left(x-5\right)}}=}$${\displaystyle \frac{x+2}{x-3}}$;

6)

${\displaystyle x^2+9x+8=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 9^2-4\cdot 1\cdot 8=}$${\displaystyle 81-32=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-9\pm \sqrt{49}}{2\cdot 1}=}$${\displaystyle \frac{-9\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-9+7}{2}=}$${\displaystyle -1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-9-7}{2}=}$${\displaystyle -8}$;

${\displaystyle x^2+9x+8=}$${\displaystyle (x+1)(x+8)}$;

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

${\displaystyle D=}$${\displaystyle 8^2-4\cdot (-1)\cdot 9=}$${\displaystyle 64+36=}$${\displaystyle 100}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-8\pm \sqrt{100}}{2\cdot (-1)}=}$${\displaystyle \frac{-8\pm 10}{-2}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{-8-10}{-2}=}$${\displaystyle 9}$;

${\displaystyle -x^2+8x+9=}$${\displaystyle -(x+1)(x-9)}$;

${\displaystyle \frac{x^2+9x+8}{-x^2+8x+9}=}$${\displaystyle \frac{\cancel{\left(x+1\right)}(x+8)}{-\cancel{\left(x+1\right)}(x-9)}=}$${\displaystyle \frac{x+8}{-(x-9)}=}$${\displaystyle \frac{x+8}{9-x}}$.