ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 617

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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.

\[\textbf{а)}\ 3x^{2} - 24x + 21 =\]

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

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

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

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

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

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

\[\textbf{б)}\ 5z^{2} + 10z - 15 =\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\textbf{г)}\ x^{2} - 12x + 20 =\]

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

\[x² - 12x + 20 = 0\]

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

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

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

\[\textbf{д)} - y^{2} + 16y - 15 =\]

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

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

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

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

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

\[y_{2} = 8 - 7 = 1.\]

\[\textbf{е)} - t^{2} - 8t + 9 =\]

\[= - (t - 1)(t + 9)\]

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

\[t^{2} + 8t - 9 = 0\]

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

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

\[t_{2} = - 4 - 5 = - 9.\]

\[\textbf{ж)}\ 2x^{2} - 5x + 3 =\]

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

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

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

\[D = 5^{2} - 2 \cdot 3 \cdot 4 = 25 - 24 = 1\]

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

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

\[\textbf{з)}\ 5y^{2} + 2y - 3 =\]

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

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

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

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

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

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

\[\textbf{и)} - 2n^{2} + 5n + 7 =\]

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

\[= (n + 1)(7 - 2n)\]

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

\[2n^{2} - 5n - 7 = 0\]

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

\[= 81\]

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

\[n_{2} = \frac{5 - 9}{4} = - 1.\]