ГДЗ по Алгебре 8 Класс Номер 624 Макарычев Миндюк — Подробные Ответы

Сократите дробь:
a)
\[
\frac{4x + 4}{3x^2 + 2x — 1};
\]
б)
\[
\frac{2a^2 — 5a — 3}{3a — 9};
\]
в)
\[
\frac{16 — b^2}{b^2 — b — 12};
\]
г)
\[
\frac{2y^2 + 7y + 3}{y^2 — 9};
\]
д)
\[
\frac{p^2 — 11p + 10}{20 + 8p — p^2};
\]
е)
\[
\frac{3x^2 + 16x — 12}{10 — 13x — 3x^2}.
\]

a)
\[ \frac{4x + 4}{3x^2 + 2x — 1} = \frac{4(x+1)}{(x+1)(3x-1)} = \frac{4}{3x-1} \]

\[ 3x^2 + 2x — 1 = 0 \]

\[ D = k^2 — ac = 1^2 — 3 \cdot (-1) = 1 + 3 = 4 > 0 \]

\[ \sqrt{D} = 2 \]

\[ x_1 = \frac{-1 + 2}{3} = \frac{1}{3} \]

\[ x_2 = \frac{-1 — 2}{3} = \frac{-3}{3} = -1 \]

\[ 3x^2 + 2x — 1 = 3(x + 1)\left(x — \frac{1}{3}\right) = (x + 1)(3x — 1) \]

б)
\[ \frac{2a^2 — 5a — 3}{3a — 9} = \frac{(2a + 1)(a — 3)}{3(a — 3)} = \frac{2a + 1}{3} \]

\[ 2a^2 — 5a — 3 = 0 \]

\[ D = b^2 — 4ac = (-5)^2 — 4 \cdot 2 \cdot (-3) = 25 + 24 = 49 > 0 \]

\[ \sqrt{D} = 7 \]

\[ a_1 = \frac{5 + 7}{2 \cdot 2} = \frac{12}{4} = 3 \]

\[ a_2 = \frac{5 — 7}{2 \cdot 2} = \frac{-2}{4} = -0.5 \]

\[ 2a^2 — 5a — 3 = 2(a + 0.5)(a — 3) = (2a + 1)(a — 3) \]

в)
\[ \frac{16 — b^2}{b^2 — b — 12} = \frac{(4-b)(4+b)}{(b+3)(b-4)} = \frac{-(4+b)}{b+3} = \frac{-4-b}{b+3} \]

\[ b^2 — b — 12 = 0 \]

\[ D = b^2 — 4ac = (-1)^2 — 4 \cdot 1 \cdot (-12) = 1 + 48 = 49 > 0 \]

\[ \sqrt{D} = 7 \]

\[ b_1 = \frac{1 + 7}{2 \cdot 1} = \frac{8}{2} = 4 \]

\[ b_2 = \frac{1 — 7}{2 \cdot 1} = \frac{-6}{2} = -3 \]

\[ b^2 — b — 12 = (b + 3)(b — 4) \]

г)
\[ \frac{2y^2 + 7y + 3}{y^2 — 9} = \frac{(y+3)(2y+1)}{(y+3)(y-3)} = \frac{2y+1}{y-3} \]

\[ 2y^2 + 7y + 3 = 0 \]

\[ D = b^2 — 4ac = 7^2 — 4 \cdot 2 \cdot 3 = 49 — 24 = 25 > 0 \]

\[ \sqrt{D} = 5 \]

\[ y_1 = \frac{-7 + 5}{2 \cdot 2} = \frac{-2}{4} = -0.5 \]

\[ y_2 = \frac{-7 — 5}{2 \cdot 2} = \frac{-12}{4} = -3 \]

\[ 2y^2 + 7y + 3 = 2(y + 3)(y + 0.5) = (y + 3)(2y + 1) \]

д)
\[ \frac{p^2 — 11p + 10}{20 + 8p — p^2} = \frac{(p-1)(p-10)}{-(p+2)(p-10)} = \frac{1-p}{p+2} \]

\[ p^2 — 11p + 10 = 0 \]

\[ D = b^2 — 4ac = (-11)^2 — 4 \cdot 1 \cdot 10 = 121 — 40 = 81 > 0 \]

\[ \sqrt{D} = 9 \]

\[ p_1 = \frac{11 + 9}{2 \cdot 1} = \frac{20}{2} = 10 \]

\[ p_2 = \frac{11 — 9}{2 \cdot 1} = \frac{2}{2} = 1 \]

\[ p^2 — 11p + 10 = (p — 1)(p — 10) \]

\[ 20 + 8p — p^2 = 0 \]

\[ D = b^2 — 4ac = 8^2 — 4 \cdot (-1) \cdot 20 = 64 + 80 = 144 > 0 \]

\[ \sqrt{D} = 12 \]

\[ p_1 = \frac{-8 + 12}{2 \cdot (-1)} = \frac{4}{-2} = -2 \]

\[ p_2 = \frac{-8 — 12}{2 \cdot (-1)} = \frac{-20}{-2} = 10 \]

\[ 20 + 8p — p^2 = -(p + 2)(p — 10) \]

e)
\[ \frac{3x^2 + 16x — 12}{10 — 13x — 3x^2} = \frac{(3x — 2)(x + 6)}{(x + 5)(-3x + 2)} = \frac{(3x — 2)(x + 6)}{-(x + 5)(3x — 2)} = \frac{x + 6}{-x — 5} \]

Решение \( 3x^2 + 16x — 12 = 0 \):
\[ D = b^2 — 4ac = 16^2 — 4 \cdot 3 \cdot (-12) = 256 + 144 = 400 > 0 \]

\[ \sqrt{D} = 20 \]

\[ x_1 = \frac{-16 + 20}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3} \]

\[ x_2 = \frac{-16 — 20}{2 \cdot 3} = \frac{-36}{6} = -6 \]

\[ 3x^2 + 16x — 12 = 3\left(x — \frac{2}{3}\right)(x + 6) = (3x — 2)(x + 6) \]

Решение \( 10 — 13x — 3x^2 = 0 \):

\[ D = b^2 — 4ac = (-13)^2 — 4 \cdot (-3) \cdot 10 = 169 + 120 = 289 > 0 \]

\[ \sqrt{D} = 17 \]

\[ x_1 = \frac{13 + 17}{2 \cdot (-3)} = \frac{30}{-6} = -5 \]

\[ x_2 = \frac{13 — 17}{2 \cdot (-3)} = \frac{-4}{-6} = \frac{2}{3} \]

\[ 10 — 13x — 3x^2 = -3(x + 5)\left(x — \frac{2}{3}\right) = (x + 5)(-3x + 2) \]

а)