Помогите, пожалуйста, решить неравенство: 20^x+4*10^x > 5^x+1 - Все ответы

Дан 1 ответ

5^{x+1}" alt="20^x+4 \cdot 10^x > 5^{x+1}" align="absmiddle" class="latex-formula">

5 \cdot 5^x" alt="(5 \cdot 4)^x+4 \cdot (2 \cdot 5)^x > 5 \cdot 5^x" align="absmiddle" class="latex-formula">

0" alt="5^x \cdot 4^x+4 \cdot 2^x \cdot 5^x - 5 \cdot 5^x>0" align="absmiddle" class="latex-formula">

0" alt="5^x \cdot (4^x+4 \cdot 2^x-5)>0" align="absmiddle" class="latex-formula">

0" alt="5^x \cdot ((2^2)^x+4 \cdot 2^x-5)>0" align="absmiddle" class="latex-formula">

0" alt="5^x \cdot (2^{2x}+4 \cdot 2^x-5)>0" align="absmiddle" class="latex-formula">

0" alt="x \in (- \infty ;+ \infty ) \Rightarrow 5^x>0" align="absmiddle" class="latex-formula">

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0" alt="2^{2x}+4 \cdot 2^x-5>0" align="absmiddle" class="latex-formula">

0" alt="2^x=t, t>0" align="absmiddle" class="latex-formula">

0" alt="t^2+4t-5>0" align="absmiddle" class="latex-formula">

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

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

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

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

0\end{cases} \Rightarrow t \in \left( 1;+ \infty \right)" alt="\begin{cases}t \in \left( - \infty ;-5\right) \cup \left( 1;+ \infty \right)\\ t>0\end{cases} \Rightarrow t \in \left( 1;+ \infty \right)" align="absmiddle" class="latex-formula">

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

1" alt="2^x>1" align="absmiddle" class="latex-formula">

2^0" alt="2^x>2^0" align="absmiddle" class="latex-formula">

0" alt="x>0" align="absmiddle" class="latex-formula">

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

_zn (654k баллов) 22 Май, 20