Решебник по алгебре 7 класс Мерзляк ФГОС Задание 612

 

\[\boxed{\text{612\ (612).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[4a^{2} - 12ax + 9x^{2} + x^{2} - 2x + 1 =\]

\[= 10 \cdot \left( x^{2} - 4 \right)\]

\[- 2x - 12ax = - 4a^{2} - 41\]

\[- 2x \cdot (1 + 6a) = - 4a^{2} - 41\]

\[1 + 6a = 0\]

\[6a = - 1\ \ \]

\[a = - \frac{1}{6}\]

\[при\ \ a = - \frac{1}{6}\ уравнение\ \]

\[не\ имеет\ корней:\]

\[4 \cdot \left( - \frac{1}{6} \right) - 12x \cdot \left( - \frac{1}{6} \right) - 2x =\]

\[= - 41\]

\[- \frac{2}{3} + 2x - 2x = - 41\]

\[0x = - 41 + \frac{2}{3}\]

\[0 \neq - 40\frac{1}{3}\]

\[Ответ:\ при\ a = - \frac{1}{6}.\]

\[\boxed{\text{612.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ (2a)^{2} = 4a^{2}.\]

\[2)\ \left( a^{2} \right)^{2} = a^{4}.\]

\[3)\ \left( 3b^{3} \right)^{2} = 9b^{6}.\]

\[4)\ \left( 7x^{4} \right)^{2} = 49x^{8}.\]

\[5)\ (0,3x)^{2} = 0,09x^{2}.\]

\[6)\ \left( 0,4y^{5}z^{2} \right)^{2} = 0,16y^{10}z^{4}.\]

\[7)\ \left( \frac{1}{6}a^{2}b^{3}c^{4} \right)^{2} = \frac{1}{36}a^{4}b^{6}c^{8}.\]

\[8)\ \left( 1\frac{1}{3}m^{6}n \right)^{2} = \left( \frac{4}{3}m^{6}n \right)^{2} =\]

\[= \frac{16}{9}m^{12}n^{2} = 1\frac{7}{9}m^{12}n^{2}\text{.\ }\]