ГДЗ по алгебре 8 класс Макарычев Задание 426

\[\boxed{\text{426\ (426).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left( \sqrt{x} + 1 \right)\left( \sqrt{x} - 1 \right) =\]

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

\[\textbf{б)}\ \left( \sqrt{x} - \sqrt{a} \right)\left( \sqrt{x} + \sqrt{a} \right) =\]

\[= \left( \sqrt{x} \right)^{2} - \left( \sqrt{a} \right)^{2} = x - a\]

\[\textbf{в)}\ \left( \sqrt{m} + \sqrt{2} \right)^{2} =\]

\[= \left( \sqrt{m} \right)^{2} + 2\sqrt{m}\sqrt{2} + \left( \sqrt{2} \right)^{2} =\]

\[= m + 2\sqrt{2m} + 2\]

\[\textbf{г)}\ \left( \sqrt{3} - \sqrt{x} \right)^{2} =\]

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

\[= 3 - 2\sqrt{3x} + x\]

\[\textbf{д)}\ \left( 5\sqrt{7} - 13 \right)\left( 5\sqrt{7} + 13 \right) =\]

\[= \left( 5\sqrt{7} \right)^{2} - 13^{2} = 25 \cdot 7 - 169 =\]

\[= 175 - 169 = 6\]

\[\textbf{е)}\ \left( 2\sqrt{2} + 3\sqrt{3} \right)\left( 2\sqrt{2} - 3\sqrt{3} \right) =\]

\[= \left( 2\sqrt{2} \right)^{2} - 3\left( \sqrt{3} \right)^{2} =\]

\[= 4 \cdot 2 - 9 \cdot 3 = 8 - 27 = - 19\]

\[\textbf{ж)}\ \left( 6 - \sqrt{2} \right)^{2} + 3\sqrt{32} =\]

\[= 36 - 12\sqrt{2} + 2 + 3\sqrt{16 \cdot 2} =\]

\[= 38 - 12\sqrt{2} + 12\sqrt{2} = 38\]

\[\textbf{з)}\ \left( \sqrt{2} + \sqrt{18} \right)^{2} - 30 =\]

\[= 2 + 2\sqrt{2}\sqrt{18} + 18 - 30 =\]

\[= - 10 + 2\sqrt{36} =\]

\[= - 10 + 12 = 2\]