ГДЗ по алгебре 9 класс Макарычев Задание 83

\[\boxed{\text{83\ (83).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{4x + 4}{3x^{2} + 2x - 1}\ \]

\[3x^{2} + 2x - 1 = 0\]

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = \frac{- 1 - 2}{3} = - 1;\ \ x_{2} =\]

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

\[3x^{2} + 2x - 1 =\]

\[= 3 \cdot (x + 1)\left( x - \frac{1}{3} \right) =\]

\[= (x + 1)(3x - 1).\]

\[\Longrightarrow \frac{4x + 4}{3x^{2} + 2x - 1} =\]

\[= \frac{4 \cdot (x + 1)}{(x + 1)(3x - 1)} = \frac{4}{3x - 1}.\]

\[\textbf{б)}\ \frac{2a^{2} - 5a - 3}{3a - 9}\]

\[2a^{2} - 5a - 3 = 0\]

\[D = 5^{2} + 4 \cdot 2 \cdot 3 =\]

\[= 25 + 24 = 49\]

\[a_{1} = \frac{5 + 7}{4} = 3,\ \ a_{2} =\]

\[= \frac{5 - 7}{4} = - \frac{1}{2}.\]

\[2a^{2} - 5a - 3 =\]

\[= 2 \cdot (a - 3)\left( a + \frac{1}{2} \right) =\]

\[= (a - 3)(2a + 1).\]

\[\Longrightarrow \frac{2a^{2} - 5a - 3}{3a - 9} =\]

\[= \frac{(a - 3)(2a + 1)}{3 \cdot (a - 3)} = \frac{2a + 1}{3}.\]

\[\textbf{в)}\ \frac{16 - b^{2}}{b^{2} - b - 12}\]

\[b^{2} - b - 12 = 0\]

\[D = 1 + 4 \cdot 12 = 49\]

\[b_{1} = \frac{1 + 7}{2} = 4;\ \ \ b_{2} =\]

\[= \frac{1 - 7}{2} = - 3.\]

\[b^{2} - b - 12 = (b - 4)(b + 3).\]

\[\Longrightarrow \frac{16 - b^{2}}{b^{2} - b - 12} =\]

\[= \frac{(4 - b)(4 + b)}{(b - 4)(b + 3)} = - \frac{4 + b}{b + 3}.\]

\[\textbf{г)}\ \frac{2y^{2} + 7y + 3}{y^{2} - 9}\]

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

\[D = 7^{2} - 4 \cdot 2 \cdot 3 =\]

\[= 49 - 24 = 25;\]

\[y_{1} = \frac{- 7 - 5}{4} = - 3;\ \ y_{2} =\]

\[= \frac{- 7 + 5}{4} = - \frac{1}{2}.\]

\[2y^{2} + 7y + 3 =\]

\[= 2 \cdot (y + 3)\left( y + \frac{1}{2} \right) =\]

\[= (y + 3)(2y + 1).\]

\[\Longrightarrow \frac{2y^{2} + 7y + 3}{y^{2} - 9} =\]

\[= \frac{(y + 3)(2y + 1)}{(y - 3)(y + 3)} = \frac{2y + 1}{y - 3}.\]

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

\[= \frac{p^{2} - 11p + 10}{- \left( p^{2} - 8p - 20 \right)}\]

\[p^{2} - 11p + 10 = 0\]

\[D = 11^{2} - 4 \cdot 10 =\]

\[= 121 - 40 = 81\]

\[p_{1} = \frac{11 + 9}{2} = 10;\ \ \ p_{2} =\]

\[= \frac{11 - 9}{2} = 1.\]

\[p^{2} - 11p + 10 =\]

\[= (p - 10)(p - 1).\]

\[p^{2} - 8p - 20 = 0\]

\[D_{1} = 4^{2} + 20 = 36\]

\[p_{1} = 4 + 6 = 10;\ \ \ p_{2} =\]

\[= 4 - 6 = - 2.\]

\[p^{2} - 8p - 20 = (p + 2)(p - 10).\]

\[\Longrightarrow \frac{p^{2} - 11p + 10}{- \left( p^{2} - 8p - 20 \right)} =\]

\[= \frac{(p - 10)(p - 1)}{- (p - 10)(p + 2)} = \frac{1 - p}{p + 2}.\]

\[\textbf{е)}\ \frac{3x^{2} + 16x - 12}{10 - 13x - 3x^{2}} =\]

\[= \frac{3x^{2} + 16x - 12}{- \left( 3x^{2} + 13x - 10 \right)}\]

\[3x^{2} + 16x - 12 = 0\]

\[D_{1} = 8^{2} + 3 \cdot 12 = 100\]

\[x_{1} = \frac{- 8 - 10}{3} = - 6;\ \ \ x_{2} =\]

\[= \frac{- 8 + 10}{3} = \frac{2}{3}.\]

\[3x^{2} + 16x - 12 =\]

\[= 3 \cdot \left( x - \frac{2}{3} \right)(x + 6) =\]

\[= (x + 6)(3x - 2).\]

\[3x^{2} + 13x - 10 = 0\]

\[D = 13^{2} + 4 \cdot 3 \cdot 10 = 289\]

\[x_{1} = \frac{- 13 + 17}{6} = \frac{2}{3};\ \ x_{2} =\]

\[= \frac{- 13 - 17}{6} = - 5.\]

\[3x^{2} + 13x - 10 =\]

\[= 3 \cdot \left( x - \frac{2}{3} \right)(x + 5) =\]

\[= (x + 5)(3x - 2).\]

\[\Longrightarrow \frac{3x^{2} + 16x - 12}{- \left( 3x^{2} + 13x - 10 \right)} =\]

\[= \frac{(x + 6)(3x - 2)}{- (x + 5)(3x - 2)} = - \frac{x + 6}{x + 5}.\]