Решебник по алгебре 7 класс Мерзляк ФГОС Задание 562

 

\[\boxed{\text{562\ (562).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{2x - 1}{8} - \frac{x + 2}{4} = x\ \ \ \ \ | \cdot 8\]

\[2x - 1 - 2 \cdot (x + 2) = 8 \cdot x\]

\[2x - 1 - 2x - 4 = 8x\]

\[- 5 = 8x\]

\[x = - \frac{5}{8}\]

\[Ответ:\ x = - \frac{5}{8}.\]

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

\[= - 1\]

\[6x + 9 - 6x - 10 = - 1\]

\[0 \cdot x = 0\]

\[Ответ:бесконечно\ много\ \]

\[корней.\]

\[\boxed{\text{562.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[= 4^{2} \cdot (x - y)^{2} = 16 \cdot (x - y)^{2}.\]

\[2)\ (18a + 27b)^{2} =\]

\[= (9 \cdot 2a + 9 \cdot 3b)^{2} =\]

\[= \left( 9 \cdot (2a + 3b) \right)^{2} =\]

\[= 9^{2} \cdot (2a + 3b)^{2} =\]

\[= 81 \cdot (2a + 3b)^{2}.\]

\[3)\ (8m - 10n)^{3} =\]

\[= (2 \cdot 4m - 2 \cdot 5n)^{3} =\]

\[= \left( 2 \cdot (4m - 5n) \right)^{3} =\]

\[= 2^{3} \cdot (4m - 5n)^{3} =\]

\[= 8 \cdot (4m - 5n)^{3}.\]

\[4)\ \left( a^{2} - 9a \right)^{2} = \left( a \cdot (a - 9) \right)^{2} =\]

\[= a^{2} \cdot (a - 9)^{2}.\]

\[5)\ \left( 16x^{2}y + 40xy^{2} \right)^{2} =\]

\[= \left( 8xy \cdot (2x + 5y) \right)^{2} =\]

\[= (8xy)^{2} \cdot (4x + 10y)^{2} =\]

\[= 64x^{2}y^{2} \cdot (2x + 5y)^{2}.\]

\[6)\ \left( 22x^{4} - 28x^{2}y^{3} \right)^{5} =\]

\[= \left( 2x^{2} \cdot 11x^{2} - 2x^{2} \cdot 14y^{3} \right)^{5} =\]

\[= (2x^{2} \cdot {\left( 11x^{2} - 14y^{3} \right))}^{5}\ =\]

\[= \left( 2x^{2} \right)^{5} \cdot \left( 11x^{2} - 14y^{3} \right)^{5} =\]

\[= 32x^{10} \cdot \left( 11x^{2} - 14y^{3} \right)^{5}\text{.\ }\]