ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 827

\[\boxed{\text{827.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

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

\[+ 25 = x^{4} - 10x^{2} + 25\]

\[\textbf{б)}\ \left( 7 - y^{3} \right)^{2} = 49 - 2 \cdot 7y^{3} +\]

\[+ \left( y^{3} \right)^{2} = 49 - 14y^{3} + y^{6}\]

\[\textbf{в)}\ \left( 2a + b^{4} \right)^{2} = (2a)^{2} + 4ab^{4} +\]

\[+ \left( b^{4} \right)^{2} = 4a² + 4ab^{4} + b^{8}\]

\[\textbf{г)}\ \left( - 3p + q^{3} \right)² = ( - 3p)^{2} -\]

\[- 6pq^{3} + \left( q^{3} \right)^{2} = 9p^{2} -\]

\[- 6pq^{3} + q^{6}\ \]

\[\boxed{\text{828.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ \left( a^{2} - 3a \right)^{2} = \left( a^{2} \right)^{2} - 6a^{2}a +\]

\[+ (3a)^{2} = a^{4} - 6a^{3} + 9a²\]

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

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

\[= \frac{1}{4}x^{6} + 6x^{4} + 36x²\]

\[\textbf{в)}\ \left( c^{2} - 0,7c^{3} \right)^{2} = \left( c^{2} \right)^{2} -\]

\[- 2 \cdot 0,7c^{2}c^{3} + \left( 0,7c^{3} \right)^{2} = c^{4} -\]

\[- 1,4c^{5} + 0,49c^{6}\]

\[\textbf{г)}\ \left( 4y^{3} - 0,5y^{2} \right)^{2} = \left( 4y^{3} \right)^{2} -\]

\[- 2 \cdot 4 \cdot 0,5y^{3}y^{2} + \left( 0,5y^{2} \right)^{2} =\]

\[= 16y^{6} - 4y^{5} + 0,25y^{4}\]

\[\textbf{д)}\ \left( 1\frac{1}{2}a^{5} + 8a^{2} \right)^{2} =\]

\[= \left( \frac{3}{2}a^{5} + 8a^{2} \right)^{2} = \left( \frac{3}{2}a^{5} \right)^{2} +\]

\[+ 2 \cdot 8 \cdot \frac{3}{2}a^{5} \cdot a² + \left( 8a^{2} \right)^{2} =\]

\[= \frac{9}{4}a^{10} + 24a^{7} + 64a^{4}\]

\[\textbf{е)}\ \left( 0,6b - 60b^{2} \right)^{2} = (0,6b)^{2} -\]

\[- 2 \cdot 0,6 \cdot b \cdot b^{2} + \left( 60b^{2} \right)^{2} =\]

\[= 0,36b^{2} - 72b^{3} + 3600b^{4}\]