Question

1.вычислить : [33*(4^1/4)^-12+2^-5/-2]^-1
2.решить неравенство sqrt(3x^3+1)>2sqrt(x)
3.решить уравнение 2*5^sqrt x+25*5^sqrt x=135

Answer 1

$1)\; \; \; \Big (33\cdot (4^{1/4})^{-12}+\frac{2^{-5}}{-2}\Big)^{-1}=\Big (33\cdot 4^{-3}-2^{-4}\Big )^{-1}=\\\\=\Big (\frac{33}{64}-\frac{1}{16}\Big )^{-1}=\Big ( \frac{33-4}{64} \Big )^{-1}= \frac{64}{29} =2 \frac{6}{29} \\\\2)\; \; \; \sqrt{3x^3+1} \ \textgreater \ 2 \sqrt{x} \\\\\left \{ {{2\sqrt{x}\geq 0} \atop {3x^3+1\ \textgreater \ 4x}} \right. \; \; \left \{ {{x\geq 0} \atop {3x^3-4x+1\ \textgreater \ 0}} \right. \; \; \left \{ {{x\geq 0} \atop {(x-1)(3x^2+3x-1)\ \textgreater \ 0}} \right. \\\\3x^2+3x-1=0\; ,\; \; D=9+4\cdot 3=21$

$x_1= \frac{-3-\sqrt{21}}{6}\approx -1,26\; \; ,\; \; x_2= \frac{-3+\sqrt{21}}{6}\approx 0,26\\\\ (x-1)\cdot (x- \frac{-3-\sqrt{21}}{6})\cdot (x- \frac{-3+\sqrt{21}}{6} )\ \textgreater \ 0\\\\Znaki:\; \; ---( \frac{-3-\sqrt{21}}{6} )+++(-1)---( \frac{-3+\sqrt{21}}{6} )+++\\\\x\in ( \frac{-3-\sqrt{21}}{6} ;-1)\cup ( \frac{-3+\sqrt{21}}{6};+\infty )\\\\Tak\; kak \; \; x \geq 0\; ,\; \; to\; \; otvet:\; \; x\in ( \frac{-3+\sqrt{21}}{6},+\infty )\; .$

$3)\; \; \; 2\cdot 5^{ \sqrt{x} }+25\cdot 5^{ \sqrt{x} }=135\\\\27\cdot 5^{ \sqrt{x} }=135\\\\5^{\sqrt{x} }=5\\\\ \sqrt{x} =1\\\\x=1$