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$1)\,\,{\log _{0,5}}({9^{x - 1}} + 1) - 2 \ge {\log _{0,5}}({3^{x - 1}} + 7)$
Đặt $t = {3^{x - 1}} > 0$
Bpt $
\Leftrightarrow {\log _{0,5}}({t^2} + 1) > 2 + {\log _{0,5}}(t + 7)\\
\Leftrightarrow \log_{0,5}({t^2} + 1) > {\log _{0,5}}(\frac{{t + 7}}{4})\\
\Leftrightarrow {t^2} + 1 < \frac{{t + 7}}{4}\\
\Leftrightarrow 4{t^2} - t - 3 < 0 \Leftrightarrow 0 < t < 1\\
\Leftrightarrow {3^{x - 1}} < 1 \Rightarrow x < 1
$
$\begin{array}{l}
2)\,\,\sqrt {{x^2} - 2x + 5} + \sqrt {x - 1} = 2\\
\Leftrightarrow \sqrt {{{(x - 1)}^2} + 4} + \sqrt {x - 1} = 2\\
\sqrt {{{(x - 1)}^2} + 4} \ge 2\\
\sqrt {x - 1} \ge 0\\
\Rightarrow VT \ge 2
\end{array}$
Dấu bằng xảy ra khi $x = 1.$
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