ГДЗ по алгебре 9 класс Макарычев Задание 220

\[\boxed{\text{220\ (220).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 2x^{2} - 3x + 7 =\]

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

\[= 2 \cdot \left( x^{2} - 2 \cdot \frac{3}{4}x + \frac{9}{16} - \frac{9}{16} + \frac{7}{2} \right) =\]

\[= 2 \cdot \left( \left( x - \frac{3}{4} \right)^{2} - \frac{9}{16} + \frac{56}{16} \right) =\]

\[= 2 \cdot \left( \left( x - \frac{3}{4} \right)^{2} + \frac{47}{16} \right) =\]

\[= 2 \cdot \left( x - \frac{3}{4} \right)^{2} + \frac{47}{8} =\]

\[= 2 \cdot \left( x - \frac{3}{4} \right)^{2} + 5\frac{7}{8}\]

\[\textbf{б)} - 3x^{2} + 4x - 1 =\]

\[= - 3 \cdot \left( x^{2} - 2 \cdot \frac{2}{3}x + \frac{4}{9} - \frac{4}{9} + \frac{1}{3} \right) =\]

\[= - 3 \cdot \left( \left( x - \frac{2}{3} \right)^{2} - \frac{1}{9} \right) =\]

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

\[\textbf{в)}\ 5x^{2} - 3x =\]

\[= 5 \cdot \left( x^{2}\ - 2 \cdot \frac{3}{10}x + \frac{9}{100} - \frac{9}{100} \right) =\]

\[= 5 \cdot \left( \left( x - \frac{3}{10} \right)^{2} - \frac{9}{100} \right) =\]

\[= 5 \cdot \left( x - \frac{3}{10} \right)^{2} - \frac{9}{20}\]

\[\textbf{г)} - 4x^{2} + 8x =\]

\[= - 4 \cdot \left( x^{2} - 2x + 1 - 1 \right) =\]

\[= - 4 \cdot \left( (x - 1)^{2} - 1 \right) =\]

\[= - 4 \cdot (x - 1)^{2} + 4\]