763 Вправа 763Алгебра 8 клас Тарасенкова 2025 · § 18 · стор. 227

763. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x^2+3x}{2}=}$${\displaystyle \frac{7+x}{4}}$ | ${\displaystyle \cdot 4}$

${\displaystyle \frac{(x^2+3x)\cdot \stackrel{2}{\cancel{4}}}{\underset{1}{\cancel{2}}}=}$${\displaystyle \frac{(7+x)\cdot \cancel{4}}{\cancel{4}}}$;

${\displaystyle 2(x^2+3x)=}$${\displaystyle 7+x}$;

${\displaystyle 2x^2+6x-x-7=}$${\displaystyle 0}$;

${\displaystyle 2x^2+5x-7=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 5^2-4\cdot 2\cdot (-7)=}$${\displaystyle 25+56=}$${\displaystyle 81}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm \sqrt{81}}{2\cdot 2}=}$${\displaystyle \frac{-5\pm 9}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+9}{4}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-9}{4}=}$${\displaystyle \frac{-14}{4}=}$${\displaystyle -3,5}$;

2) ${\displaystyle \frac{x^2}{5}-\frac{2x}{3}=}$${\displaystyle \frac{5+x}{6}}$ | ${\displaystyle \cdot 30}$

${\displaystyle \frac{x^2\cdot \stackrel{6}{\cancel{30}}}{\underset{1}{\cancel{5}}}-\frac{2x\cdot \stackrel{10}{\cancel{30}}}{\underset{1}{\cancel{3}}}=}$${\displaystyle \frac{(5+x)\cdot \stackrel{5}{\cancel{30}}}{\underset{1}{\cancel{6}}}}$;

${\displaystyle 6x^2-10\cdot 2x=}$${\displaystyle 5(5+x)}$;

${\displaystyle 6x^2-20x=}$${\displaystyle 5(5+x)}$;

${\displaystyle 6x^2-20x-25-5x=}$${\displaystyle 0}$;

${\displaystyle 6x^2-25x-25=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-25)}^2-4\cdot 6\cdot (-25)=}$${\displaystyle 625+600=}$${\displaystyle 1225}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{25\pm \sqrt{1225}}{2\cdot 6}=}$${\displaystyle \frac{25\pm 35}{12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{25+35}{12}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{25-35}{12}=}$${\displaystyle \frac{-10}{12}=}$${\displaystyle -\frac{5}{6}}$;

3) ${\displaystyle \frac{y+y^2}{2}-\frac{y^2-6}{6}=}$${\displaystyle \frac{1-3y}{4}}$ | ${\displaystyle \cdot 12}$

${\displaystyle \frac{(y+y^2)\cdot \stackrel{6}{\cancel{12}}}{\underset{1}{\cancel{2}}}-\frac{(y^2-6)\cdot \stackrel{2}{\cancel{12}}}{\underset{1}{\cancel{6}}}=}$${\displaystyle \frac{(1-3y)\cdot \stackrel{3}{\cancel{12}}}{\underset{1}{\cancel{4}}}}$;

${\displaystyle 6(y+y^2)-2(y^2-6)=}$${\displaystyle 3(1-3y)}$;

${\displaystyle 6y+6y^2-2y^2+12=}$${\displaystyle 3-9y}$;

${\displaystyle 4y^2+15y+9=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {15}^2-4\cdot 4\cdot 9=}$${\displaystyle 225-144=}$${\displaystyle 81}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-15\pm \sqrt{81}}{2\cdot 4}=}$${\displaystyle \frac{-15\pm 9}{8}}$;

${\displaystyle y_1=}$${\displaystyle \frac{-15+9}{8}=}$${\displaystyle \frac{-6}{8}=}$${\displaystyle -0,75}$;

${\displaystyle y_2=}$${\displaystyle \frac{-15-9}{8}=}$${\displaystyle -3}$;

4) ${\displaystyle \frac{y(y-7)}{3}-y=}$${\displaystyle \frac{4-y}{3}-4}$ | ${\displaystyle \cdot 3}$

${\displaystyle \frac{y(y-7)\cdot \cancel{3}}{\cancel{3}}-y\cdot 3=}$${\displaystyle \frac{(4-y)\cdot \cancel{3}}{\cancel{3}}-4\cdot 3}$;

${\displaystyle y(y-7)-3y=}$${\displaystyle 4-y-12}$;

${\displaystyle y^2-7y-3y=}$${\displaystyle 4-y-12}$;

${\displaystyle y^2-10y+y+8=}$${\displaystyle 0}$;

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

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

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

${\displaystyle y_1=}$${\displaystyle \frac{9+7}{2}=}$${\displaystyle 8}$;

${\displaystyle y_2=}$${\displaystyle \frac{9-7}{2}=}$${\displaystyle 1}$.