ГДЗ по алгебре 9 класс Макарычев Задание 76

\[\boxed{\text{76\ (76).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 3x^{2} - 24x + 21 = 0\]

\[3 \cdot \left( x^{2} - 8x + 7 \right) = 0\]

\[x^{2} - 8x + 7 = 0\]

\[D_{1} = 4^{2} - 7 = 16 - 7 = 9\]

\[x_{1} = 4 + 3 = 7;\ \ x_{2} = 4 - 3 = 1.\]

\[\Longrightarrow 3x^{2} - 24x + 21 =\]

\[= 3 \cdot (x - 7)(x - 1).\]

\[\textbf{б)}\ 5x^{2} + 10x - 15 = 0\]

\[5 \cdot \left( x^{2} + 2x - 3 \right) = 0\]

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

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

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

\[= - 1 - 2 = - 3.\]

\[\Longrightarrow 5x^{2} + 10x - 15 =\]

\[= 5 \cdot (x + 3)(x - 1).\]

\[\textbf{в)}\ \frac{1}{6}x^{2} + \frac{1}{2}x + \frac{1}{3} = 0\]

\[\frac{1}{6} \cdot \left( x^{2} + 3x + 2 \right) = 0\]

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

\[D = 9 - 4 \cdot 2 = 9 - 8 = 1\]

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

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

\[\Longrightarrow \frac{1}{6}x^{2} + \frac{1}{2}x + \frac{1}{3} =\]

\[= \frac{1}{6} \cdot (x + 2)(x + 1).\]

\[\textbf{г)}\ x^{2} - 12x + 20 = 0\]

\[D_{1} = 6^{2} - 20 = 36 - 20 = 16\]

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

\[= 6 - 4 = 2.\]

\[\Longrightarrow x^{2} - 12x + 20 =\]

\[= (x - 2)(x - 10).\]

\[\textbf{д)} - y^{2} + 16y - 15 = 0\]

\[- \left( y^{2} - 16y + 15 \right) = 0\]

\[D_{1} = 8^{2} - 15 = 64 - 15 = 49\]

\[y_{1} = 8 + 7 = 15;\ \ \ y_{2} =\]

\[= 8 - 7 = 1.\]

\[\Longrightarrow - y^{2} + 16y - 15 =\]

\[= - (y - 1)(y - 15).\]

\[\textbf{е)} - x^{2} - 8x + 9 = 0\]

\[- \left( x^{2} + 8x - 9 \right) = 0\]

\[D_{1} = 4^{2} + 9 = 16 + 9 = 25\]

\[x_{1} = - 4 + 5 = 1;\ \ \ x_{2} =\]

\[= - 4 - 5 = - 9.\]

\[\Longrightarrow - x^{2} - 8x + 9 =\]

\[= - (x - 1)(x + 9).\]

\[\textbf{ж)}\ 2x^{2} - 5x + 3 = 0\]

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

\[= 25 - 24 = 1\]

\[x_{1} = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2};\ \ \ x_{2} =\]

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

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

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

\[= (2x - 3)(x - 1).\]

\[\textbf{з)}\ 5y^{2} + 2y - 3 = 0\]

\[D_{1} = 1 + 3 \cdot 5 = 16\]

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

\[= \frac{- 1 + 4}{5} = \frac{3}{5}.\]

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

\[= 5 \cdot \left( y - \frac{3}{5} \right)(y + 1) =\]

\[= (5y - 3)(y + 1).\]

\[\textbf{и)} - 2x^{2} + 5x + 7 = 0\]

\[- \left( 2x^{2} - 5x - 7 \right) = 0\]

\[D = 5^{2} + 4 \cdot 2 \cdot 7 = 25 + 56 = 81\]

\[x_{1} = \frac{5 + 9}{4} = \frac{14}{4} = \frac{7}{2};\ \ x_{2} =\]

\[= \frac{5 - 9}{4} = - 1.\]

\[\Longrightarrow - 2x^{2} + 5x + 7 =\]

\[= - 2 \cdot \left( x - \frac{7}{2} \right)(x + 1) =\]

\[= (x + 1)(7 - 2x)\text{.\ }\]