Question

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Answer 1

$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 + 3}}{{\sqrt x + 2}}\\
= 1 + \dfrac{3}{{\sqrt x + 2}}\\
Do:\sqrt x + 2 > 0\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 2}} > 0\\
\Leftrightarrow 1 + \dfrac{3}{{\sqrt x + 2}} > 1\\
\Leftrightarrow K > 1 > 0\\
\Leftrightarrow K - 1 > 0\\
\Leftrightarrow K.\left( {K - 1} \right) > 0\\
\Leftrightarrow {K^2} - K > 0\\
\Leftrightarrow {K^2} > K\\
2)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 + 4}}{{\sqrt x + 1}}\\
= 1 + \dfrac{4}{{\sqrt x + 1}} > 1 > 0\\
\Leftrightarrow \sqrt K > 1\\
\Leftrightarrow \sqrt K - 1 > 0\\
\Leftrightarrow \sqrt K .\left( {\sqrt K - 1} \right) > 0\\
\Leftrightarrow K - \sqrt K > 0\\
\Leftrightarrow K > \sqrt K
\end{array}$$\begin{array}{l}
3)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{x + 4}}{{\sqrt x + 1}}\\
= \dfrac{{x - 1 + 5}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 5}}{{\sqrt x + 1}}\\
= \sqrt x - 1 + \dfrac{5}{{\sqrt x + 1}}\\
= \sqrt x + 1 + \dfrac{5}{{\sqrt x + 1}} - 2\\
Do:\sqrt x + 1 > 0\\
Theo\,Co - si:\\
\sqrt x + 1 + \dfrac{5}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{5}{{\sqrt x + 1}}} = 2\sqrt 5 \\
\Leftrightarrow K \ge 2\sqrt 5 - 2 > 1\\
\Leftrightarrow {K^{2020}} > K\\
4)\\
Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{ - \sqrt x - 4}}{{\sqrt x + 2}}\\
= - \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}\\
= - \left( {\dfrac{{\sqrt x + 2 + 2}}{{\sqrt x + 2}}} \right)\\
= - \left( {1 + \dfrac{2}{{\sqrt x + 2}}} \right)\\
Do:\dfrac{2}{{\sqrt x + 2}} > 0\\
\Leftrightarrow 1 + \dfrac{2}{{\sqrt x + 2}} > 1\\
\Leftrightarrow - \left( {1 + \dfrac{2}{{\sqrt x + 2}}} \right) < - 1 < 0\\
\Leftrightarrow K < 0\\
\Leftrightarrow \left| K \right| > K
\end{array}$

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Answer 2

$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 + 3}}{{\sqrt x + 2}}\\
= 1 + \dfrac{3}{{\sqrt x + 2}}\\
Do:\sqrt x + 2 > 0\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 2}} > 0\\
\Leftrightarrow 1 + \dfrac{3}{{\sqrt x + 2}} > 1\\
\Leftrightarrow K > 1 > 0\\
\Leftrightarrow K - 1 > 0\\
\Leftrightarrow K.\left( {K - 1} \right) > 0\\
\Leftrightarrow {K^2} - K > 0\\
\Leftrightarrow {K^2} > K\\
2)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 + 4}}{{\sqrt x + 1}}\\
= 1 + \dfrac{4}{{\sqrt x + 1}} > 1 > 0\\
\Leftrightarrow \sqrt K > 1\\
\Leftrightarrow \sqrt K - 1 > 0\\
\Leftrightarrow \sqrt K .\left( {\sqrt K - 1} \right) > 0\\
\Leftrightarrow K - \sqrt K > 0\\
\Leftrightarrow K > \sqrt K
\end{array}$$\begin{array}{l}
3)Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{x + 4}}{{\sqrt x + 1}}\\
= \dfrac{{x - 1 + 5}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 5}}{{\sqrt x + 1}}\\
= \sqrt x - 1 + \dfrac{5}{{\sqrt x + 1}}\\
= \sqrt x + 1 + \dfrac{5}{{\sqrt x + 1}} - 2\\
Do:\sqrt x + 1 > 0\\
Theo\,Co - si:\\
\sqrt x + 1 + \dfrac{5}{{\sqrt x + 1}} \ge 2\sqrt {\left( {\sqrt x + 1} \right).\dfrac{5}{{\sqrt x + 1}}} = 2\sqrt 5 \\
\Leftrightarrow K \ge 2\sqrt 5 - 2 > 1\\
\Leftrightarrow {K^{2020}} > K\\
4)\\
Dkxd:x \ge 0;x \ne 4\\
K = \dfrac{{ - \sqrt x - 4}}{{\sqrt x + 2}}\\
= - \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}\\
= - \left( {\dfrac{{\sqrt x + 2 + 2}}{{\sqrt x + 2}}} \right)\\
= - \left( {1 + \dfrac{2}{{\sqrt x + 2}}} \right)\\
Do:\dfrac{2}{{\sqrt x + 2}} > 0\\
\Leftrightarrow 1 + \dfrac{2}{{\sqrt x + 2}} > 1\\
\Leftrightarrow - \left( {1 + \dfrac{2}{{\sqrt x + 2}}} \right) < - 1 < 0\\
\Leftrightarrow K < 0\\
\Leftrightarrow \left| K \right| > K
\end{array}$

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