Question

70 балів! Спростити вираз!!! ПОТРІБНО НЕГАЙНО!!!​

Answer 1

$\displaystyle\bf\\\frac{\sqrt[4]{a} }{\sqrt[4]{a}-1 } -\frac{\sqrt[4]{a} }{\sqrt[4]{a} +1} \\\\ODZ \ :\\\\\\\left\{\begin{array}{ccc}a\geq 0\\\sqrt[4]{a} \neq -1\\\sqrt[4]{a} \neq 1\end{array}\right\\\\\\\frac{\sqrt[4]{a} \cdot(\sqrt[4]{a} +1)-\sqrt[4]{a} \cdot(\sqrt[4]{a} -1)}{(\sqrt[4]{a}-1)\cdot(\sqrt[4]{a}+1) } =\frac{\sqrt{a}+\sqrt[4]{a} -\sqrt{a} +\sqrt[4]{a} }{(\sqrt[4]{a} )^{2} -1^{2} } =\\\\\\=\frac{2\sqrt[4]{a} }{\sqrt{a} -1}$

10)

$\displaystyle\bf\\5^{2x+1} -7\cdot 10^{x} +2^{2x+1} > 0\\\\\\5^{2x} \cdot 5-7\cdot 10^{x} +2^{2x} \cdot 2 > 0\\\\\\5\cdot 25^{x} -7\cdot 10^{x} +2\cdot 4^{x} > 0 \ |:4^{x} > 0\\\\\\5\cdot\frac{25^{x} }{4^{x} } -7\cdot\frac{10^{x} }{4^{x} } +2\cdot\frac{4^{x} }{4^{x} } > 0\\\\\\5\cdot\Big(\frac{5}{2} \Big)^{2x} -7\cdot\Big(\frac{5}{2} \Big)^{x} +2 > 0\\\\\\\Big(\frac{5}{2} \Big)^{x} =m > 0\\\\\\5m^{2}-7m+2 > 0\\\\\\5\cdot(m-0,4)\cdot(m-1) > 0\\\\\\+ + + + + (0,4)- - - - - (1)+ + + + +$

$\displaystyle\bf\\\left \{ {{m < 0,4} \atop {m > 1}} \right. \\\\\\1)\\\\m < 0,4\\\\m < \frac{2}{5} \\\\\\\Big(\frac{5}{2} \Big)^{x} < \frac{2}{5} \\\\\\\Big(\frac{5}{2} \Big)^{x} < \Big(\frac{5}{2} \Big)^{-1} \\\\\\x < -1\\\\2)\\\\m > 1\\\\\\\Big(\frac{5}{2} \Big)^{x} > 1\\\\\\\Big(\frac{5}{2} \Big)^{x} > \Big(\frac{5}{2} \Big)^{0} \\\\\\x > 0\\\\\\Otvet \ : \ x\in(-\infty \ ; \ -1)\cup(0 \ ; \ +\infty)$

Answer 2

Ответ:

ответ смотри на фотографии, дай пожалуйста корону