ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1095

\[\boxed{\text{1095.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ (2x - 3y)^{2} + (2x + 3y)^{2} =\]

\[= 4x^{2} - 12xy + 9y^{2} + 4x^{2} +\]

\[+ 12xy + 9y^{2} =\]

\[= 8x² + 18y²\]

\[\textbf{б)}\ (2x + 3y)^{2} - (2x - 3y)^{2} =\]

\[= 4x^{2} + 12xy + 9y^{2} -\]

\[- \left( 4x^{2} - 12xy + 9y^{2} \right) =\]

\[= 4x^{2} + 12xy + 9y^{2} - 4x^{2} +\]

\[+ 12xy - 9y^{2} = 24xy\]

\[\textbf{в)}\ 2 \cdot \left( \frac{x}{2} + \frac{y}{4} \right)^{2} + (2x - y)^{2} =\]

\[= 2 \cdot \left( \frac{x^{2}}{4} + \frac{\text{xy}}{4} + \frac{y^{2}}{16} \right) + 4x^{2} -\]

\[- 4xy + y^{2} =\]

\[= \frac{x^{2}}{2} + \frac{\text{xy}}{2} + \frac{y^{2}}{8} + 4x^{2} - 4xy +\]

\[+ y^{2} = 4,5x^{2} - 3,5xy + 1,125y²\]

\[\textbf{г)}\ 3 \cdot \left( \frac{x}{3} + \frac{y}{9} \right)^{2} - (3x - y)^{2} =\]

\[= 3 \cdot \left( \frac{x^{2}}{9} + \frac{2xy}{27} + \frac{y^{2}}{81} \right) -\]

\[- \left( 9x^{2} - 6xy + y^{2} \right) =\]

\[= \frac{x^{2}}{3} + \frac{2xy}{9} + \frac{y^{2}}{27} - 9x^{2} +\]

\[+ 6xy - y^{2} =\]

\[= \frac{x^{2} - 27x^{2}}{3} + \frac{2xy + 54xy}{9} +\]

\[+ \frac{y^{2} - 27y^{2}}{27} = \ - \frac{26}{3}x^{2} +\]

\[+ \frac{56}{9}xy - \frac{26}{27}y²\]