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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 755
Задание 755
\[\boxed{\mathbf{755\ (755).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[a,\ b,\ c;\ \ \ \ \ то\ b = \frac{a + c}{2};\ \ \]
\[2b = a + c,\ \]
\[a + c = \left( \sqrt{a} \right)^{2} + \left( \sqrt{c} \right)^{2} =\]
\[= \left( \sqrt{a} \right)^{2} + \left( \sqrt{c} \right)^{2} + 2\sqrt{\text{ac}} -\]
\[- 2\sqrt{\text{ac}} = \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}},\]
\[то\ есть:\ \ \ \]
\[2b = \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}} \Longrightarrow \ \]
\[\Longrightarrow \ b = \frac{\left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}}}{2}\]
\[\frac{1}{\sqrt{a} + \sqrt{b}} + \frac{1}{\sqrt{b} + \sqrt{c}} = \frac{2}{\sqrt{a} + \sqrt{c}}\]
\[\frac{1}{\sqrt{a} + \sqrt{b}} + \frac{1}{\sqrt{b} + \sqrt{c}} =\]
\[= \frac{\sqrt{b} + \sqrt{c} + \sqrt{a} + \sqrt{b}}{\left( \sqrt{a} + \sqrt{b} \right)\left( \sqrt{b} + \sqrt{c} \right)} =\]
\[= \frac{\sqrt{a} + \sqrt{c} + 2\sqrt{b}}{\left( \sqrt{a} + \sqrt{b} \right)\left( \sqrt{b} + \sqrt{c} \right)} =\]
\[= \frac{\sqrt{a} + \sqrt{c} + 2\sqrt{b}}{\sqrt{\text{ab}} + \sqrt{\text{ac}} + b + \sqrt{\text{bc}}} =\]
\[= \frac{2 \cdot \left( \sqrt{a} + \sqrt{c} + 2\sqrt{b} \right)}{2\sqrt{\text{ab}} + 2\sqrt{\text{ac}} + 2b + 2\sqrt{\text{bc}}} =\]
\[= \frac{2 \cdot \left( \sqrt{a} + \sqrt{c} + 2\sqrt{b} \right)}{2\sqrt{\text{ab}} + 2\sqrt{\text{ac}} + \left( \sqrt{a} + \sqrt{c} \right)^{2} - 2\sqrt{\text{ac}} + 2\sqrt{\text{bc}}} =\]
\[= \frac{2 \cdot \left( \sqrt{a} + \sqrt{c} + 2\sqrt{b} \right)}{\left( \sqrt{a} + \sqrt{c} \right)\left( 2\sqrt{b} + \sqrt{a} + \sqrt{c} \right)} =\]
\[= \frac{2}{\sqrt{a} + \sqrt{c}} \Longrightarrow доказано.\]
\[\boxed{\mathbf{755.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ 5x² - 11x + 2 \leq 0\]
\[5x^{2} - 11x + 2 = 0\]
\[D = 121 - 40 = 81\]
\[x_{1} = \frac{11 + 9}{10} = 2;\ \ \]
\[x_{2} = \frac{11 - 9}{10} = \frac{1}{5} = 0,2\]
\[Ответ:x \in \lbrack 0,2;2\rbrack.\]
\[2)\ 4x² + 3,6x > 0\]
\[4x(x + 0,9) > 0\]
\[x = 0;\ \ x = - 0,9\]
\[Ответ:x\]
\[\in ( - \infty; - 0,9) \cup (0;\ + \infty).\]
\[3)\ 12 - 5x - 3x^{2} \leq 0\ \ \ | \cdot ( - 1)\]
\[3x^{2} + 5x - 12 \geq 0\]
\[D = 25 + 144 = 169\]
\[x_{1} = \frac{- 5 + 13}{6} = \frac{4}{3} = 1\frac{1}{3};\ \ \ \ \]
\[\ x_{2} = \frac{- 5 - 13}{6} = - 3\]
\[Ответ:\]
\[x \in ( - \infty;\ - 3\rbrack \cup \left\lbrack 1\frac{1}{3};\ + \infty \right).\]
\[4)\ 0,04 - x^{2} > 0\]
\[(0,2 - x)(0,2 + x) > 0\]
\[x < 0,2;\ \ \ x > - 0,2\]
\[Ответ:( - 0,2;0,2).\]
