1087 Вправа 1087Алгебра 8 клас Бевз 2025 · § 21 · стор. 204

1087. Скороти дріб. а) (x² + 2x)/(x² + 7x + 10); б) (x² − 4x + 3)/(x² − 9); в) (x² + 3x + 2)/(x² + 4x + 3); г) (3c² − 5c + 2)/(3c² − c − 2); ґ) (x² + 2x − 15)/(35 + 2x − x²); д) (−3x² + 5x − 2)/(x² − 1); е) (2a² − 5a + 2)/(3a² − 3,5a − 5); є) (a² + 9a + 14)/(a² + 10a + 21); ж) (c² + √5c − 10)/(c² − 3√5c + 10).

Розкладаємо чисельник і знаменник на множники і скорочуємо спільний множник.

а) ${\displaystyle \frac{x^2+2x}{x^2+7x+10}}$;

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

${\displaystyle x^2+7x+10}$: ${\displaystyle x_1+x_2=}$${\displaystyle -7}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 10}$ → ${\displaystyle x_1=}$${\displaystyle -2,x_2=}$${\displaystyle -5}$;

${\displaystyle x^2+7x+10=}$${\displaystyle \left(x+2\right)\left(x+5\right)}$;

${\displaystyle \frac{x\cancel{\left(x+2\right)}}{\cancel{\left(x+2\right)}(x+5)}=}$${\displaystyle \frac{x}{x+5}}$;

б) ${\displaystyle \frac{x^2-4x+3}{x^2-9}}$;

${\displaystyle x^2-4x+3}$: ${\displaystyle x_1+x_2=}$${\displaystyle 4}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 3}$ → ${\displaystyle x_1=}$${\displaystyle 3,x_2=}$${\displaystyle 1}$;

${\displaystyle x^2-4x+3=}$${\displaystyle \left(x-3\right)\left(x-1\right)}$;

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

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

в) ${\displaystyle \frac{x^2+3x+2}{x^2+4x+3}}$;

${\displaystyle x^2+3x+2}$: ${\displaystyle x_1+x_2=}$${\displaystyle -3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 2}$ → ${\displaystyle x_1=}$${\displaystyle -1,x_2=}$${\displaystyle -2}$;

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

${\displaystyle x^2+4x+3}$: ${\displaystyle x_1+x_2=}$${\displaystyle -4}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 3}$ → ${\displaystyle x_1=}$${\displaystyle -1,x_2=}$${\displaystyle -3}$;

${\displaystyle x^2+4x+3=}$${\displaystyle \left(x+1\right)\left(x+3\right)}$;

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

г) ${\displaystyle \frac{3c^2-5c+2}{3c^2-c-2}}$;

${\displaystyle 3c^2-5c+2}$: ${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 3\cdot 2=}$${\displaystyle 1}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{5\pm 1}{6}}$;

${\displaystyle c_1=}$${\displaystyle 1}$;

${\displaystyle c_2=}$${\displaystyle \frac{2}{3}}$;

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

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

${\displaystyle c_{1,2}=}$${\displaystyle \frac{1\pm 5}{6}}$;

${\displaystyle c_1=}$${\displaystyle 1}$;

${\displaystyle c_2=}$${\displaystyle -\frac{2}{3}}$;

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

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

ґ) ${\displaystyle \frac{x^2+2x-15}{35+2x-x^2}}$;

${\displaystyle x^2+2x-15}$: ${\displaystyle x_1+x_2=}$${\displaystyle -2}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -15}$ → ${\displaystyle x_1=}$${\displaystyle 3,x_2=}$${\displaystyle -5}$;

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

${\displaystyle -x^2+2x+35}$: ${\displaystyle D=}$${\displaystyle 2^2-4\cdot (-1)\cdot 35=}$${\displaystyle 144}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-2\pm 12}{-2}}$;

${\displaystyle x_1=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle -5}$;

${\displaystyle -x^2+2x+35=}$${\displaystyle -\left(x-7\right)\left(x+5\right)}$;

${\displaystyle \frac{(x-3)\cancel{\left(x+5\right)}}{-(x-7)\cancel{\left(x+5\right)}}=}$${\displaystyle -\frac{x-3}{x-7}=}$${\displaystyle \frac{3-x}{x-7}}$;

д) ${\displaystyle \frac{-3x^2+5x-2}{x^2-1}}$;

${\displaystyle -3x^2+5x-2}$: ${\displaystyle D=}$${\displaystyle 5^2-4\cdot (-3)\cdot (-2)=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-5\pm 1}{-6}}$;

${\displaystyle x_1=}$${\displaystyle 1}$;

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

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

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

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

е) ${\displaystyle \frac{2a^2-5a+2}{3a^2-3,5a-5}}$;

${\displaystyle 2a^2-5a+2}$: ${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 2=}$${\displaystyle 9}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{5\pm 3}{4}}$;

${\displaystyle a_1=}$${\displaystyle 2}$;

${\displaystyle a_2=}$${\displaystyle 0,5}$;

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

${\displaystyle 3a^2-3,5a-5=}$${\displaystyle \frac{1}{2}(6a^2-7a-10)}$;

${\displaystyle 6a^2-7a-10}$: ${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 6\cdot (-10)=}$${\displaystyle 289}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{7\pm 17}{12}}$;

${\displaystyle a_1=}$${\displaystyle 2}$;

${\displaystyle a_2=}$${\displaystyle -\frac{5}{6}}$;

${\displaystyle 6a^2-7a-10=}$${\displaystyle 6(a-2)(a+\frac{5}{6})=}$${\displaystyle \left(a-2\right)\left(6a+5\right)}$;

${\displaystyle \frac{(2a-1)\cancel{\left(a-2\right)}}{\frac{1}{2}\cancel{\left(a-2\right)}(6a+5)}=}$${\displaystyle \frac{2(2a-1)}{6a+5}=}$${\displaystyle \frac{4a-2}{6a+5}}$;

є) ${\displaystyle \frac{a^2+9a+14}{a^2+10a+21}}$;

${\displaystyle a^2+9a+14}$: ${\displaystyle a_1+a_2=}$${\displaystyle -9}$;

${\displaystyle a_1{a}_2=}$${\displaystyle 14}$ → ${\displaystyle a_1=}$${\displaystyle -2,a_2=}$${\displaystyle -7}$;

${\displaystyle a^2+9a+14=}$${\displaystyle \left(a+2\right)\left(a+7\right)}$;

${\displaystyle a^2+10a+21}$: ${\displaystyle a_1+a_2=}$${\displaystyle -10}$;

${\displaystyle a_1{a}_2=}$${\displaystyle 21}$ → ${\displaystyle a_1=}$${\displaystyle -3,a_2=}$${\displaystyle -7}$;

${\displaystyle a^2+10a+21=}$${\displaystyle \left(a+3\right)\left(a+7\right)}$;

${\displaystyle \frac{(a+2)\cancel{\left(a+7\right)}}{(a+3)\cancel{\left(a+7\right)}}=}$${\displaystyle \frac{a+2}{a+3}}$;

ж) ${\displaystyle \frac{c^2+\sqrt{5}c-10}{c^2-3\sqrt{5}c+10}}$;

чисельник: ${\displaystyle D=}$${\displaystyle 5+40=}$${\displaystyle 45}$;

${\displaystyle \sqrt{45}=}$${\displaystyle 3\sqrt{5}}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{-\sqrt{5}\pm 3\sqrt{5}}{2}}$: ${\displaystyle \sqrt{5}}$ і ${\displaystyle -2\sqrt{5}}$;

знаменник: ${\displaystyle D=}$${\displaystyle 45-40=}$${\displaystyle 5}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{3\sqrt{5}\pm \sqrt{5}}{2}}$: ${\displaystyle 2\sqrt{5}}$ і ${\displaystyle \sqrt{5}}$;

${\displaystyle \frac{\cancel{\left(c-\sqrt{5}\right)}(c+2\sqrt{5})}{(c-2\sqrt{5})\cancel{\left(c-\sqrt{5}\right)}}=}$${\displaystyle \frac{c+2\sqrt{5}}{c-2\sqrt{5}}}$.