ГДЗ по алгебре 8 класс Макарычев Задание 423

\[\boxed{\text{423\ (423).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

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

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

\[\textbf{в)}\ \left( \sqrt{11} - 3 \right)\left( \sqrt{11} + 3 \right) =\]

\[= \left( \sqrt{11} \right)^{2} - 3^{2} = 11 - 9 = 2\]

\[\textbf{г)}\ \left( \sqrt{10} + \sqrt{7} \right)\left( \sqrt{7} - \sqrt{10} \right) =\]

\[= \left( \sqrt{7} \right)^{2} - \left( \sqrt{10} \right)^{2} = 7 - 10 =\]

\[= - 3\]

\[\textbf{д)}\ \left( \sqrt{a} + \sqrt{b} \right)^{2} =\]

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

\[= a + 2\sqrt{\text{ab}} + b\]

\[\textbf{е)}\ \left( \sqrt{m} - \sqrt{n} \right)^{2} =\]

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

\[= m - 2\sqrt{\text{mn}} + n\]

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

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

\(= 2 + 6\sqrt{2} + 9 = 11 + 6\sqrt{2}\)

\[\textbf{з)}\ \left( \sqrt{5} - \sqrt{2} \right)^{2} =\]

\[= \left( \sqrt{5} \right)^{2} - 2\sqrt{5}\sqrt{2} + \left( \sqrt{2} \right)^{2} =\]

\[= 5 - 2\sqrt{10} + 2 = 7 - 2\sqrt{10}\]