1088 Вправа 1088Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1088. Скороти дріб. а) (3x − 9)/(2x² − 5x − 3); б) (a² − 9)/(2a² + 7a + 3); в) (c² − 8c − 20)/(c² − 11c + 10); г) (a² − 8a + 7)/(a² − 9a + 14); ґ) (2 − 3c + c²)/(c² − 4c + 4); д) (2x² − 3x + 1)/(3x² − 4x + 1); е) (x² + 2x − 15)/(x² + 4x − 5); є) (x² − 3)/(x² − 2√3x + 3); ж) (2c² − 5c − 3)/(2c² + 7c + 3).

а) ${\displaystyle \frac{3x-9}{2x^2-5x-3}}$;

${\displaystyle 2x^2-5x-3}$: ${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot (-3)=}$${\displaystyle 49}$;

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

${\displaystyle x_1=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle -0,5}$;

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

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

б) ${\displaystyle \frac{a^2-9}{2a^2+7a+3}}$;

${\displaystyle 2a^2+7a+3}$: ${\displaystyle D=}$${\displaystyle 7^2-4\cdot 2\cdot 3=}$${\displaystyle 25}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{-7\pm 5}{4}}$;

${\displaystyle a_1=}$${\displaystyle -0,5}$;

${\displaystyle a_2=}$${\displaystyle -3}$;

${\displaystyle 2a^2+7a+3=}$${\displaystyle 2(a+0,5)(a+3)=}$${\displaystyle \left(2a+1\right)\left(a+3\right)}$;

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

в) ${\displaystyle \frac{c^2-8c-20}{c^2-11c+10}}$;

${\displaystyle c^2-8c-20}$: ${\displaystyle c_1+c_2=}$${\displaystyle 8}$;

${\displaystyle c_1{c}_2=}$${\displaystyle -20}$ → ${\displaystyle c_1=}$${\displaystyle 10,c_2=}$${\displaystyle -2}$;

${\displaystyle c^2-8c-20=}$${\displaystyle \left(c-10\right)\left(c+2\right)}$;

${\displaystyle c^2-11c+10}$: ${\displaystyle c_1+c_2=}$${\displaystyle 11}$;

${\displaystyle c_1{c}_2=}$${\displaystyle 10}$ → ${\displaystyle c_1=}$${\displaystyle 10,c_2=}$${\displaystyle 1}$;

${\displaystyle c^2-11c+10=}$${\displaystyle \left(c-10\right)\left(c-1\right)}$;

${\displaystyle \frac{\cancel{\left(c-10\right)}(c+2)}{(c-1)\cancel{\left(c-10\right)}}=}$${\displaystyle \frac{c+2}{c-1}}$;

г) ${\displaystyle \frac{a^2-8a+7}{a^2-9a+14}}$;

${\displaystyle a^2-8a+7}$: ${\displaystyle a_1+a_2=}$${\displaystyle 8}$;

${\displaystyle a_1{a}_2=}$${\displaystyle 7}$ → ${\displaystyle a_1=}$${\displaystyle 7,a_2=}$${\displaystyle 1}$;

${\displaystyle a^2-8a+7=}$${\displaystyle \left(a-7\right)\left(a-1\right)}$;

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

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

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

${\displaystyle \frac{(a-1)\cancel{\left(a-7\right)}}{(a-2)\cancel{\left(a-7\right)}}=}$${\displaystyle \frac{a-1}{a-2}}$;

ґ) ${\displaystyle \frac{2-3c+c^2}{c^2-4c+4}}$;

${\displaystyle c^2-3c+2}$: ${\displaystyle c_1+c_2=}$${\displaystyle 3}$;

${\displaystyle c_1{c}_2=}$${\displaystyle 2}$ → ${\displaystyle c_1=}$${\displaystyle 2,c_2=}$${\displaystyle 1}$;

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

${\displaystyle c^2-4c+4=}$${\displaystyle {(c-2)}^2}$;

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

д) ${\displaystyle \frac{2x^2-3x+1}{3x^2-4x+1}}$;

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

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

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

${\displaystyle x_2=}$${\displaystyle 0,5}$;

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

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

${\displaystyle x_{1,2}=}$${\displaystyle \frac{4\pm 2}{6}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{1}{3}}$;

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

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

е) ${\displaystyle \frac{x^2+2x-15}{x^2+4x-5}}$;

${\displaystyle x^2+2x-15}$: ${\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)}$;

${\displaystyle x^2+4x-5}$: ${\displaystyle x_1+x_2=}$${\displaystyle -4}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -5}$ → ${\displaystyle x_1=}$${\displaystyle 1,x_2=}$${\displaystyle -5}$;

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

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

є) ${\displaystyle \frac{x^2-3}{x^2-2\sqrt{3}x+3}}$;

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

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

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

ж) ${\displaystyle \frac{2c^2-5c-3}{2c^2+7c+3}}$;

${\displaystyle 2c^2-5c-3}$: ${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot (-3)=}$${\displaystyle 49}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{5\pm 7}{4}}$;

${\displaystyle c_1=}$${\displaystyle 3}$;

${\displaystyle c_2=}$${\displaystyle -0,5}$;

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

${\displaystyle 2c^2+7c+3}$: ${\displaystyle D=}$${\displaystyle 7^2-4\cdot 2\cdot 3=}$${\displaystyle 25}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{-7\pm 5}{4}}$;

${\displaystyle c_1=}$${\displaystyle -0,5}$;

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

${\displaystyle 2c^2+7c+3=}$${\displaystyle 2(c+0,5)(c+3)=}$${\displaystyle \left(2c+1\right)\left(c+3\right)}$;

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