Question

Решить неравенство: 20^x+4*10^x > 5^x+1

Answer 1

$20^x+4 \cdot 10^x 5^{x+1}$

$(5 \cdot 4)^x+4 \cdot (2 \cdot 5)^x 5 \cdot 5^x$

$5^x \cdot 4^x+4 \cdot 2^x \cdot 5^x - 5 \cdot 5^x0$

$5^x \cdot (4^x+4 \cdot 2^x-5)0$

$5^x \cdot ((2^2)^x+4 \cdot 2^x-5)0$

$5^x \cdot (2^{2x}+4 \cdot 2^x-5)0$

$x \in (- \infty ;+ \infty ) \Rightarrow 5^x0$

----------------------

$2^{2x}+4 \cdot 2^x-50$

$2^x=t, t0$

$t^2+4t-50$

$\Delta=4^2-4 \cdot 1 \cdot (-5)=16+20=36$

$\sqrt{\Delta}= \sqrt{36}=6$

$t_1= \frac{-4-6}{2}= \frac{-10}{2}=-5$

$t_2= \frac{-4+6}{2}= \frac{2}{2}=1$

$\begin{cases}t \in \left( - \infty ;-5\right) \cup \left( 1;+ \infty \right)\\ t0\end{cases} \Rightarrow t \in \left( 1;+ \infty \right)$

----------------------

$2^x\in \left( 1;+ \infty \right)$

$2^x1$

$2^x2^0$

$x0$

$x \in \left(0;+ \infty \right)$