Решебник по алгебре 9 класс Мерзляк Задание 755

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\[a,\ b,\ c;\ \ \ \ \ то\ b = \frac{a + c}{2};\ \ \]

\[2b = a + c,\ \]

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

\[= \left( \sqrt{a} \right)^{2} + \left( \sqrt{c} \right)^{2} + 2\sqrt{\text{ac}} -\]

\[- 2\sqrt{\text{ac}} = \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}},\]

\[то\ есть:\ \ \ \]

\[2b = \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}} \Longrightarrow \ \]

\[\Longrightarrow \ b = \frac{\left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}}}{2}\]

\[\frac{1}{\sqrt{a} + \sqrt{b}} + \frac{1}{\sqrt{b} + \sqrt{c}} = \frac{2}{\sqrt{a} + \sqrt{c}}\]

\[\frac{1}{\sqrt{a} + \sqrt{b}} + \frac{1}{\sqrt{b} + \sqrt{c}} =\]

\[= \frac{\sqrt{b} + \sqrt{c} + \sqrt{a} + \sqrt{b}}{\left( \sqrt{a} + \sqrt{b} \right)\left( \sqrt{b} + \sqrt{c} \right)} =\]

\[= \frac{\sqrt{a} + \sqrt{c} + 2\sqrt{b}}{\left( \sqrt{a} + \sqrt{b} \right)\left( \sqrt{b} + \sqrt{c} \right)} =\]

\[= \frac{\sqrt{a} + \sqrt{c} + 2\sqrt{b}}{\sqrt{\text{ab}} + \sqrt{\text{ac}} + b + \sqrt{\text{bc}}} =\]

\[= \frac{2 \cdot \left( \sqrt{a} + \sqrt{c} + 2\sqrt{b} \right)}{2\sqrt{\text{ab}} + 2\sqrt{\text{ac}} + 2b + 2\sqrt{\text{bc}}} =\]

\[= \frac{2 \cdot \left( \sqrt{a} + \sqrt{c} + 2\sqrt{b} \right)}{2\sqrt{\text{ab}} + 2\sqrt{\text{ac}} + \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}} + 2\sqrt{\text{bc}}} =\]

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

\[= \frac{2}{\sqrt{a} + \sqrt{c}} \Longrightarrow доказано.\]