ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 624

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Пояснение.

Если \(x_{1}\ и\ x_{2}\) – корни квадратного трехчлена ax²+bx+c, то:

\[ax^{2} + bx + c = a\left( a - x_{1} \right)\left( x - x_{2} \right).\]

Если квадратный трехчлен не имеет корней, то его нельзя разложить на множители.

Решение.

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

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

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

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

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

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

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

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

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

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

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

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

\[D = 5^{2} + 4 \cdot 2 \cdot 3 = 25 + 24 =\]

\[= 49\]

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

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

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

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

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

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

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

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

\[b_{1} + b_{2} = 1;\ \ \ b_{1} \cdot b_{2} = - 12\]

\[b_{1} = 4;\ \ \ b_{2} = - 3.\]

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

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

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

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

\[D = 7^{2} - 4 \cdot 2 \cdot 3 = 49 - 24 =\]

\[= 25\]

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

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

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

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

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

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

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

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

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

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

\[p_{1} = 10;\ \ p_{2} = 1.\]

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

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

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

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

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

\[p_{2} = 4 - 6 = - 2.\]

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

\[= (p + 2)(p - 10).\]

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

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

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

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

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

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

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

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

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

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

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

\[x_{1} = \frac{- 13 + 17}{6} = \frac{4}{6} = \frac{2}{3};\]

\[x_{2} = \frac{- 13 - 17}{6} = - 5.\]

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

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