814 Вправа 814Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 243

814. Розкладіть на лінійні множники квадратний тричлен:

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 3=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

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

${\displaystyle x_1=}$${\displaystyle \frac{5+1}{4}=}$${\displaystyle \frac{6}{4}=}$${\displaystyle 1,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-1}{4}=}$${\displaystyle 1}$;

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

2) ${\displaystyle 5x^2+4x-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 4^2-4\cdot 5\cdot (-1)=}$${\displaystyle 16+20=}$${\displaystyle 36}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-4\pm \sqrt{36}}{2\cdot 5}=}$${\displaystyle \frac{-4\pm 6}{10}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-4+6}{10}=}$${\displaystyle \frac{2}{10}=}$${\displaystyle 0,2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-4-6}{10}=}$${\displaystyle -1}$;

${\displaystyle 5x^2+4x-1=}$${\displaystyle 5(x-0,2)(x+1)=}$${\displaystyle \left(5x-1\right)\left(x+1\right)}$;

3) ${\displaystyle -3x^2-x+14=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot (-3)\cdot 14=}$${\displaystyle 1+168=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{1\pm \sqrt{169}}{2\cdot (-3)}=}$${\displaystyle \frac{1\pm 13}{-6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+13}{-6}=}$${\displaystyle \frac{14}{-6}=}$${\displaystyle -2\frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-13}{-6}=}$${\displaystyle 2}$;

${\displaystyle -3x^2-x+14=}$${\displaystyle -3(x+\frac{7}{3})(x-2)}$;${\displaystyle -(3x+7)(x-2)}$${\displaystyle \left(3x+7\right)\left(2-x\right)}$;

4) ${\displaystyle 6x^2-5x+1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 6\cdot 1=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{1}}{2\cdot 6}=}$${\displaystyle \frac{5\pm 1}{12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+1}{12}=}$${\displaystyle \frac{6}{12}=}$${\displaystyle 0,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-1}{12}=}$${\displaystyle \frac{4}{12}=}$${\displaystyle \frac{1}{3}}$;

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

5) ${\displaystyle -9x^2+12x-4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {12}^2-4\cdot (-9)\cdot (-4)=}$${\displaystyle 144-144=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{-12}{-18}=}$${\displaystyle \frac{2}{3}}$;

${\displaystyle -9x^2+12x-4=}$${\displaystyle -9{(x-\frac{2}{3})}^2=}$${\displaystyle -{\left(3x-2\right)}^2}$;

6) ${\displaystyle 3x^2-8x+5=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-8)}^2-4\cdot 3\cdot 5=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{8\pm \sqrt{4}}{2\cdot 3}=}$${\displaystyle \frac{8\pm 2}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{8+2}{6}=}$${\displaystyle \frac{10}{6}=}$${\displaystyle 1\frac{2}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-2}{6}=}$${\displaystyle 1}$;

${\displaystyle 3x^2-8x+5=}$${\displaystyle 3(x-\frac{5}{3})(x-1)=}$${\displaystyle \left(3x-5\right)\left(x-1\right)}$;

7) ${\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}$;

${\displaystyle 4x^2-4x+1=}$${\displaystyle 4{(x-\frac{1}{2})}^2=}$${\displaystyle {\left(2x-1\right)}^2}$;

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

${\displaystyle D=}$${\displaystyle 3^2-4\cdot (-2)\cdot 2=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-3\pm \sqrt{25}}{2\cdot (-2)}=}$${\displaystyle \frac{-3\pm 5}{-4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+5}{-4}=}$${\displaystyle \frac{2}{-4}=}$${\displaystyle -0,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-5}{-4}=}$${\displaystyle 2}$;

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